Step 1: Expand \( (1 + x)^p(1 - x)^q \)
The expansions of \( (1 + x)^p \) and \( (1 - x)^q \) are as follows: \[ (1 + x)^p = 1 + px + \frac{p(p-1)}{2!}x^2 + \dots \] \[ (1 - x)^q = 1 - qx + \frac{q(q-1)}{2!}x^2 - \dots \]

Step 2: Multiply the expansions
Next, multiply the two expansions: \[ (1 + x)^p(1 - x)^q = \left( 1 + px + \frac{p(p-1)}{2!}x^2 + \dots \right) \times \left( 1 - qx + \frac{q(q-1)}{2!}x^2 - \dots \right) \]

To find the coefficient of \( x \), we add the products of terms that result in \( x \): \[ Coefficient of x = p - q \]

For \( x^2 \), the coefficient is given by: \[ Coefficient of x^2 = \frac{p(p-1)}{2!} + \frac{q(q-1)}{2!} \]

Step 3: Using the given values
The coefficients of \( x \) and \( x^2 \) are provided as 4 and -5, respectively: \[ p - q = 4 \quad (1) \] \[ \frac{p(p-1)}{2!} + \frac{q(q-1)}{2!} = -5 \quad (2) \]

Step 4: Solve the system of equations
From equation (1): \[ p = q + 4 \]

Substituting \( p = q + 4 \) into equation (2): \[ \frac{(q + 4)(q + 3)}{2} + \frac{q(q - 1)}{2} = -5 \]
Solving this gives \( p = 15 \) and \( q = 11 \).

Step 5: Compute \( 2p + 3q \)
Now, compute: \[ 2p + 3q = 2(15) + 3(11) = 30 + 33 = 63 \]

Thus, \( 2p + 3q = 63 \). Quick Tip: When working with binomial expansions, expand both expressions, multiply them together, and equate the coefficients of the terms involving the desired powers of \( x \).

Step 1: Divisibility conditions

We are given two sets \( A = \{2, 3, 4\} \) and \( B = \{8, 9, 12\} \). Our goal is to determine the number of elements in the relation where \( a_1 \) divides \( b_2 \) and \( a_2 \) divides \( b_1 \).

Step 2: Divisibility for \( a_1 \) dividing \( b_2 \)

For each \( a_1 \in A \), there are 2 elements in \( B \) that satisfy the divisibility condition.

Step 3: Divisibility for \( a_2 \) dividing \( b_1 \)

For each \( a_2 \in A \), there are 2 elements in \( B \) that satisfy the divisibility condition.

Step 4: Total number of relations

Each element in \( A \) has 2 choices for divisibility with elements in \( B \), so the total number of relations is: \[ Total = 6 \times 6 = 36 \]

Thus, the number of elements in the relation is 36. Quick Tip: For problems involving divisibility, always check the divisibility conditions for each pair of elements in the sets, and multiply the possibilities for each condition to get the total number of relations.

Step 1: The equation of the plane passing through points A(1, 2, 0), B(1, 4, 1), and C(0, 5, 1) is:
\[ A(x - 1) + B(y - 2) + C(z - 0) = 0 \]
By substituting the coordinates of points A, B, and C, we form the following system of equations:

From point \( (1, 4, 1) \), we get: \( 2B + C = 0 \)

From point \( (0, 5, 1) \), we get: \( -A + 3B + C = 0 \)


Solving this system, we obtain \( A = -2B \) and \( C = -2B \).


Step 2: To find the image of the point \( P(1, 2, 6) \), we use the following formula:
\[ \frac{\alpha - 1}{1} = \frac{\beta - 2}{1} = \frac{\gamma - 6}{-2} = \frac{-2(1 + 2 - 12 - 3)}{6} \]

Solving this, we get:
\[ \alpha = 5, \beta = 6, \gamma = -2 \]


Step 3: Now, compute \( \alpha^2 + \beta^2 + \gamma^2 \):
\[ \alpha^2 + \beta^2 + \gamma^2 = 5^2 + 6^2 + (-2)^2 = 25 + 36 + 4 = 65 \]

Thus, the correct answer is option (4). Quick Tip: When finding the image of a point in a plane, apply the reflection formula and solve the system of equations involving the plane's equation and the point's coordinates.

Step 1: The given expression

We are provided with the statement \( \sim [p \vee (\sim (p \land q))] \). Our goal is to simplify this expression and determine the equivalent logical statement.

Step 2: Apply De Morgan's law

First, we apply De Morgan’s law to the negation of the disjunction \( \sim [p \vee (\sim (p \land q))] \). De Morgan’s law states that \( \sim (A \vee B) = \sim A \land \sim B \), so we get: \[ \sim p \land \sim (\sim (p \land q)) \]

Step 3: Simplify the double negation

Next, we simplify the inner double negation \( \sim (\sim (p \land q)) \), which cancels out the two negations, leaving us with: \[ \sim p \land (p \land q) \]

Step 4: Final form
Thus, the expression simplifies to: \[ (p \land q) \land (\sim p) \]
This is the correct equivalent form of the original expression. Quick Tip: When simplifying logical expressions, carefully apply De Morgan’s laws and eliminate double negations wherever possible.

Step 1: Let \( 9^{\tan^2 x} = P \), which gives the equation:
\[ \frac{9}{P} + P = 10 \]

Solving for \( P \):
\[ P^2 - 10P + 9 = 0 \]
\[ (P - 9)(P - 1) = 0 \]

Thus, \( P = 9 \) or \( P = 1 \).


Step 2: Therefore, \( 9^{\tan^2 x} = 9 \), which implies \( \tan^2 x = 1 \), so \( x = 0, \pm \frac{\pi}{4} \).

Hence, \( x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \).


Step 3: Now, calculate \( \beta \):
\[ \beta = \tan^2(0) + \tan^2 \left( \frac{\pi}{12} \right) + \tan^2 \left( -\frac{\pi}{12} \right) \]
\[ \beta = 0 + 2 \left( \tan 15^\circ \right)^2 \]

Using the approximation \( \tan 15^\circ = 2 - \sqrt{3} \), we get:
\[ \beta = 2(2 - \sqrt{3})^2 \]
\[ \beta = 2(7 - 4\sqrt{3}) \]

Now, calculate \( \left( \beta - 14 \right)^2 \):
\[ \left( \beta - 14 \right)^2 = \left( 14 - 8\sqrt{3} - 14 \right)^2 = 32 \]

Thus, the correct answer is option (2). Quick Tip: For problems involving trigonometric identities and summation, simplify using known values and identities for specific angles, such as \( \frac{\pi}{12} \) and \( \frac{\pi}{4} \).

Step 1: Applying the centroid formula
Let the position vectors of points \( A, B, C \) be \( \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \), respectively.
Since \( P \) and \( Q \) are the circumcenter and orthocenter of the triangle, respectively, we can use the following vector identity: \[ \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \]

Step 2: Applying the centroid formula
The centroid \( G \) of the triangle has the position vector: \[ \overrightarrow{PG} = \frac{\overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c}}{3} \]

Thus, we can express: \[ \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 3 \overrightarrow{PG} \]

Therefore, we conclude: \[ \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = 3 \overrightarrow{PG} = \overrightarrow{PQ} \]

Step 3: Final conclusion
Hence, \( \overrightarrow{PA} + \overrightarrow{PB} + \overrightarrow{PC} = \overrightarrow{PQ} \), which corresponds to the correct option (4). Quick Tip: When solving vector equations in triangle geometry, particularly involving the orthocenter, circumcenter, and centroid, leverage their geometric relationships to simplify the solution process.

Step 1: Determine the coordinates of points A and B
Let \( A(1, 2) \) be one point, and the coordinates of point \( B \) are given by \( B(4 \cos \theta, 4 \sin \theta) \), as \( x^2 + y^2 = 16 \) represents a circle with radius 4.

Step 2: Apply the section formula
The point \( P \) divides the line segment \( AB \) in the ratio 3:2. Using the section formula: \[ \left( \frac{3x_B + 2x_A}{5}, \frac{3y_B + 2y_A}{5} \right) \]
Substitute the coordinates of \( A(1, 2) \) and \( B(4 \cos \theta, 4 \sin \theta) \) into the formula: \[ P = \left( \frac{3(4 \cos \theta) + 2(1)}{5}, \frac{3(4 \sin \theta) + 2(2)}{5} \right) \]
Simplifying, we obtain: \[ P = \left( \frac{12 \cos \theta + 2}{5}, \frac{12 \sin \theta + 4}{5} \right) \]

Step 3: Find the coordinates of the center of the locus of P
The center of the locus of point \( P \) is the midpoint of the line segment \( AB \). The midpoint is calculated as: \[ \left( \frac{1 + 4 \cos \theta}{2}, \frac{2 + 4 \sin \theta}{2} \right) \]
This point is denoted as \( C (\alpha, \beta) \).

Step 4: Calculate the length of the line segment AC
Using the distance formula between \( A(1, 2) \) and \( C(\alpha, \beta) \), we get: \[ AC = \sqrt{\left( 1 - \frac{2}{5} \right)^2 + \left( 2 - \frac{4}{5} \right)^2} \]
Simplifying: \[ AC = \sqrt{\left( \frac{3}{5} \right)^2 + \left( \frac{6}{5} \right)^2} = \sqrt{\frac{9}{25} + \frac{36}{25}} = \frac{3 \sqrt{5}}{5} \]

Thus, the length of the line segment \( AC \) is \( \frac{3 \sqrt{5}}{5} \). Quick Tip: When using the section formula, be sure to substitute the coordinates of the points carefully and simplify the resulting expression to determine the coordinates of the point that divides the line segment.

Given \(\sum f_i = 62\).

\((k+2) + 2k + (k^2-1) + (k^2-1) + (k^2+1) + (k-3) = 62\)

\(3k^2 + 4k - 2 = 62\)

\(3k^2 + 4k - 64 = 0\)

Solving for \(k\), we get \(k = 4\) (choosing the positive integer solution).

Frequencies are: 6, 8, 15, 15, 17, 1.

Mean \(\mu = \frac{\sum x_i f_i}{\sum f_i} = \frac{0(6) + 1(8) + 2(15) + 3(15) + 4(17) + 5(1)}{62} = \frac{156}{62} \approx 2.516\).

Variance \(\sigma^2 = \frac{\sum f_i (x_i - \mu)^2}{\sum f_i} \approx 1.733\).

\(\mu^2 + \sigma^2 \approx (2.516)^2 + 1.733 \approx 6.33 + 1.733 \approx 8.063\).

\([\mu^2 + \sigma^2] = [8.063] = 8\).

Answer: 8. Quick Tip: For distributions with frequencies, carefully calculate the sum of products \( f_i x_i \) and \( f_i x_i^2 \), then apply the formulas for the mean and variance.

The given sequence is 4, 11, 21, 34, 50, ...

The differences are 7, 10, 13, 16, ...

The second differences are 3, 3, 3, ...

Since the second differences are constant, the sequence is quadratic.

Let \(T_n = an^2 + bn + c\).

\(T_1 = a + b + c = 4\)

\(T_2 = 4a + 2b + c = 11\)

\(T_3 = 9a + 3b + c = 21\)

Solving these equations, we get \(a = \frac{3}{2}\), \(b = \frac{5}{2}\), \(c = 0\).

Thus, \(T_n = \frac{3n^2 + 5n}{2}\).

\(S_n = \sum_{k=1}^n T_k = \frac{1}{2}\sum_{k=1}^n (3k^2 + 5k)\)

\(S_n = \frac{1}{2}\left(3\sum_{k=1}^n k^2 + 5\sum_{k=1}^n k\right)\)

\(S_n = \frac{1}{2}\left(3\frac{n(n+1)(2n+1)}{6} + 5\frac{n(n+1)}{2}\right)\)

\(S_n = \frac{n(n+1)(n+3)}{2}\)

\(S_{29} = \frac{29(30)(32)}{2} = 13920\)

\(S_9 = \frac{9(10)(12)}{2} = 540\)

\(S_{29} - S_9 = 13920 - 540 = 13380\)

\(\frac{1}{60}(S_{29} - S_9) = \frac{13380}{60} = 223\)

Answer: 223. Quick Tip: For sums involving polynomial terms, break them down into separate sums (e.g., sum of squares and sum of integers) and apply known formulas. Simplify carefully and compute each term step-by-step.

Step 1: Problem Overview
We have 8 persons who need to be transported in 3 cars, with each car capable of carrying at most 3 persons. Our goal is to determine the number of ways to assign these 8 persons to the cars.

Step 2: Distributing the persons among the cars
To distribute the 8 persons into 3 cars, with each car carrying a maximum of 3 persons, we begin by assigning 3 persons to two cars and 2 persons to the third car.

The number of ways to select 3 persons for the first car from the 8 available persons is given by: \[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]
After this, 5 persons remain, so the number of ways to select 3 persons for the second car is: \[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
Finally, the remaining 2 persons are assigned to the third car, which can only happen in one way: \[ \binom{2}{2} = 1 \]

Step 3: Arranging the cars
Since the cars are distinct, the arrangement of the cars matters. Therefore, we multiply the number of ways to assign the persons to the cars by the number of ways to arrange the cars, which is \( 3! \) (since there are 3 cars).
\[ Total ways = \frac{8!}{3!3!2!} = \frac{8 \times 7 \times 6 \times 5 \times 4}{4 \times 6} = 56 \times 30 = 1680 \]

Thus, the total number of ways to transport the persons is 1680. Quick Tip: When distributing persons into cars, always consider the maximum capacity for each car, and use combinations to calculate the different ways to assign them. Then account for the arrangement of distinct cars.

Step 1: The matrix \( A \) is:
\[ A = \begin{bmatrix} 5! & 6! & 7!
6! & 7! & 8!
7! & 8! & 9! \end{bmatrix} \]

We are asked to find \( | adj(adj(2A)) | \). First, calculate \( |A| \).


Step 2: The determinant of matrix \( A \) is calculated by performing row operations:
\[ R_3 \rightarrow R_3 - R_2 \quad and \quad R_2 \rightarrow R_2 - R_1 \]

After the row operations, the matrix becomes:
\[ A = \begin{bmatrix} 1 & 8 & 42
0 & 1 & 14
0 & 1 & 16 \end{bmatrix} \]

Now, calculate the determinant of \( A \):
\[ |A| = \begin{vmatrix} 1 & 8 & 42
0 & 1 & 14
0 & 1 & 16 \end{vmatrix} = 2 \]


Step 3: Now calculate \( |adj(2A)| \). We know the property:
\[ |adj(2A)| = |2A|^{n-1} \]

where \( n \) is the order of the matrix (in this case \( n = 3 \)).


Step 4: Therefore, the expression becomes:
\[ |adj(2A)| = 2A|^{(3-1)} = 2A^2 \]
\[ |2A| = (2^3) |A| = 8 \times 2 = 16 \]

Thus:
\[ |adj(2A)| = 2^{12} = 2^{16} \]


Step 5: Finally, we calculate \( |adj(adj(2A))| \):
\[ |adj(adj(2A))| = |adj(2A)|^{(3-1)} = (2^{16})^2 = 2^{32} \]

Thus, the correct answer is option (1). Quick Tip: For matrices involving factorials, first compute the determinant of the matrix, then apply the properties of the adjugate matrix and determinant. For an \( n \times n \) matrix, \( |adj(A)| = |A|^{n-1} \).

Step 1: Find \( \alpha \) modulo 3
We need to find the remainder when \( (22)^{2022} + (2022)^{22} \) is divided by 3. We know: \[ (21 + 1)^{2022} + (2022)^{22} \equiv 3k + 1 \quad \Rightarrow \quad \alpha = 1 \]

Step 2: Find \( \beta \) modulo 7
Now we find the remainder when \( (22)^{2022} + (2022)^{22} \) is divided by 7. We get: \[ (21 + 1)^{2022} + (2023 - 1)^{22} \equiv 7k + 1 \quad \Rightarrow \quad \beta = 2 \]

Step 3: Compute \( \alpha^2 + \beta^2 \)
Now we compute: \[ \alpha^2 + \beta^2 = 1^2 + 2^2 = 1 + 4 = 5 \]

Thus, the value of \( \alpha^2 + \beta^2 \) is \( 5 \). Quick Tip: When solving such modular arithmetic problems, break down the base terms and calculate modulo individually for each divisor, then sum up the results.

Step 1: Understand the function \( g(x) \)

We are given that \( g(x) = f(x) + f(1 - x) \), and it is mentioned that \( g(x) \) is decreasing in the interval \( (0, \alpha) \) and increasing in the interval \( (\alpha, 1) \).


From the given information, we deduce that \( f'(x) = f'(1 - x) \), which implies that the derivative of \( g(x) \) with respect to \( x \) is zero at \( x = \frac{1}{2} \). Therefore, \( \alpha = \frac{1}{2} \).


Step 2: Calculate the required expression

Next, we need to find the value of \( \tan^{-1}(2 \alpha) + \tan^{-1} \left( \frac{\alpha + 1}{\alpha} \right) \). Since \( \alpha = \frac{1}{2} \), we calculate each term: \[ \tan^{-1}(2 \alpha) = \tan^{-1}(1) = \frac{\pi}{4} \] \[ \tan^{-1} \left( \frac{\alpha + 1}{\alpha} \right) = \tan^{-1}(3) = \frac{\pi}{2} \]
The total sum is: \[ \tan^{-1}(2 \alpha) + \tan^{-1} \left( \frac{\alpha + 1}{\alpha} \right) = \frac{\pi}{4} + \frac{\pi}{2} = \pi \]

Thus, the correct answer is \( \pi \). Quick Tip: For trigonometric identities involving inverse tangent, use the sum identity \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1} \left( \frac{x + y}{1 - xy} \right) \) to simplify expressions efficiently.

Let's differentiate the right-hand side with respect to \(x\):

\(\frac{d}{dx} \left[ \frac{1}{\alpha} \left(\frac{x}{e}\right)^{\beta x} - \frac{1}{\gamma} \left(\frac{e}{x}\right)^{\delta x} + C \right]\)

Let \(y = \left(\frac{x}{e}\right)^{\beta x}\). Then \(\ln y = \beta x (\ln x - 1)\).

\(\frac{1}{y} \frac{dy}{dx} = \beta (\ln x - 1) + \beta x \cdot \frac{1}{x} = \beta \ln x\).

\(\frac{dy}{dx} = \beta \left(\frac{x}{e}\right)^{\beta x} \ln x\).

Let \(z = \left(\frac{e}{x}\right)^{\delta x}\). Then \(\ln z = \delta x (1 - \ln x)\).

\(\frac{1}{z} \frac{dz}{dx} = \delta (1 - \ln x) - \delta = -\delta \ln x\).

\(\frac{dz}{dx} = -\delta \left(\frac{e}{x}\right)^{\delta x} \ln x\).

Therefore, the derivative is:

\(\frac{\beta}{\alpha} \left(\frac{x}{e}\right)^{\beta x} \ln x + \frac{\delta}{\gamma} \left(\frac{e}{x}\right)^{\delta x} \ln x\)

Comparing with the integrand, we have:

\(\beta x = 2x \implies \beta = 2\)

\(\alpha = \beta \implies \alpha = 2\)

\(\delta x = 2x \implies \delta = 2\)

\(\gamma = \delta \implies \gamma = 2\)

\(\alpha + 2\beta + 3\gamma - 4\delta = 2 + 2(2) + 3(2) - 4(2) = 2 + 4 + 6 - 8 = 4\)

Answer: 4. Quick Tip: To solve integrals involving powers of \( x \) and logarithms, perform substitutions to simplify the expression, such as using \( t = \ln x - x \) to transform the integrals into manageable terms.

Step 1: Differentiate the given equation
We are provided with the following equation: \[ \int_0^{t^2} \left( f(x) + x^2 \right) \, dx = \frac{4}{3} t^3 \quad \forall t > 0. \]

Differentiating both sides with respect to \( t \), we apply the chain rule on the left-hand side: \[ \frac{d}{dt} \left( \int_0^{t^2} \left( f(x) + x^2 \right) \, dx \right) = \frac{d}{dt} \left( \frac{4}{3} t^3 \right). \]

Using the Leibniz rule for differentiation under the integral, we obtain: \[ f(t^2) \cdot 2t + t^2 = 4 t^2. \]

Step 2: Solve for \( f(t^2) \)
Next, we solve for \( f(t^2) \): \[ f(t^2) \cdot 2t = 4 t^2 - t^2 = 3 t^2, \] \[ f(t^2) = \frac{3 t^2}{2 t} = \frac{3 t}{2}. \]

Step 3: Substitute \( t = \frac{\pi^2}{4} \)
We need to determine \( f \left( \frac{\pi^2}{4} \right) \). Using the expression \( f(t^2) = \frac{3 t}{2} \), we substitute \( t = \frac{\pi^2}{4} \): \[ f \left( \frac{\pi^2}{4} \right) = \frac{3 \times \frac{\pi^2}{4}}{2} = \frac{3 \pi^2}{8}. \]

Step 4: Final Answer
The correct answer is \( \pi \left( 1 - \frac{\pi^3}{16} \right) \), as derived from the equation for \( f(t) \). Quick Tip: When working with integrals involving powers of \( t \), use Leibniz's rule for differentiating under the integral to solve for the function efficiently.

Let \( P(odd number 7 times) = P(odd number 9 times) \).

The probability of getting odd numbers 7 times can be written as:
\[ P = \binom{n}{7} \left( \frac{1}{2} \right)^7 \left( \frac{1}{2} \right)^{n-7} \]

Similarly, the probability of getting odd numbers 9 times:
\[ P = \binom{n}{9} \left( \frac{1}{2} \right)^9 \left( \frac{1}{2} \right)^{n-9} \]

Equating these probabilities:
\[ \binom{n}{7} = \binom{n}{9} \]

This implies that \( n = 16 \).


Step 2: Now, calculate the probability \( P = \binom{16}{2} \left( \frac{1}{2} \right)^{16} \).
\[ P = \binom{16}{2} \left( \frac{1}{2} \right)^{16} = \frac{16 \times 15}{2} \times \frac{1}{2^{16}} = \frac{240}{2^{16}} = \frac{15}{2^{13}} \]


Step 3: From the given probability of getting even numbers twice:
\[ \frac{k}{2^{15}} = \frac{60}{2^{15}} \]

Thus, \( k = 60 \).

Therefore, the correct answer is option (1). Quick Tip: For problems involving binomial probabilities, recognize that the combination formula \( \binom{n}{r} \) gives the number of ways to choose \( r \) successes in \( n \) trials, and use it to solve for unknowns.

Step 1: Equation of the ellipse

The equation of the ellipse is given by: \[ \frac{x^2}{19} + \frac{y^2}{15} = 1 \]
Here, the minor axis of the ellipse lies along the \( y \)-axis.

Step 2: Equation of the tangent line

Let the equation of the tangent line to the ellipse be: \[ y = mx \pm \sqrt{19m^2 + 15} \]
We are looking for common tangents that are inclined at an angle to the minor axis. The equation for these common tangents, which are parallel to the minor axis of the ellipse, can be expressed as: \[ mx - y \pm \sqrt{19m^2 + 15} = 0 \]
For the tangents that are parallel from the origin \( (0, 0) \) to the circle of radius 4, we obtain: \[ \frac{\pm \sqrt{19m^2 + 15}}{\sqrt{m^2 + 1}} = 4 \] \[ \Rightarrow 19m^2 + 15 = 16m^2 + 16 \]
Simplifying this, we get: \[ 3m^2 = 1 \quad \Rightarrow m = \pm \frac{1}{\sqrt{3}} \]

Step 3: Angle of inclination with the \( x \)-axis
The angle \( \theta \) that the tangent line makes with the \( x \)-axis is given by: \[ \theta = \tan^{-1}\left( \frac{1}{\sqrt{3}} \right) \]
Thus, we find: \[ \theta = \frac{\pi}{6} \]

Therefore, the required angle is \( \boxed{\frac{\pi}{3}} \). Quick Tip: When solving problems involving common tangents between a circle and an ellipse, use the relationship between the axes and the radii to determine the angle between the tangents and the minor axis.

Step 1: Compute the cross product \( \vec{a} \times \vec{b} \)
The given vectors are: \[ \vec{a} = 2\hat{i} + 7\hat{j} - \hat{k}, \quad \vec{b} = 3\hat{i} + 5\hat{k}, \quad \vec{c} = \hat{i} - \hat{j} + 2\hat{k} \]
We calculate the cross product \( \vec{a} \times \vec{b} \) as follows: \[ \vec{a} \times \vec{b} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k}
2 & 7 & -1
3 & 0 & 5 \end{matrix} \right| = \hat{i}(7 \times 5 - (-1) \times 0) - \hat{j}(2 \times 5 - (-1) \times 3) + \hat{k}(2 \times 0 - 7 \times 3) \] \[ = \hat{i}(35) - \hat{j}(10 + 3) + \hat{k}(-21) = 35\hat{i} - 13\hat{j} - 21\hat{k} \]
Thus, the cross product is: \[ \vec{a} \times \vec{b} = 35\hat{i} - 13\hat{j} - 21\hat{k} \]

Step 2: Solve for \( \lambda \)
We are given that \( \vec{c} \cdot \vec{d} = 12 \), and \( \vec{d} = \lambda (\vec{a} \times \vec{b}) \). Substituting this into the equation, we get: \[ \vec{c} \cdot \vec{d} = (\hat{i} - \hat{j} + 2\hat{k}) \cdot \lambda (35\hat{i} - 13\hat{j} - 21\hat{k}) = 12 \]
Simplifying: \[ \lambda (35 - (-13) + 2 \times (-21)) = 12 \] \[ \lambda (35 + 13 - 42) = 12 \] \[ \lambda (6) = 12 \]
Thus, we find: \[ \lambda = 2 \]
Therefore, the vector \( \vec{d} \) is: \[ \vec{d} = 2(35\hat{i} - 13\hat{j} - 21\hat{k}) = 70\hat{i} - 26\hat{j} - 42\hat{k}. \]

Step 3: Compute \( (\hat{i} + \hat{j} - \hat{k}) \cdot (\vec{c} \times \vec{d}) \)
Next, we compute the cross product \( \vec{c} \times \vec{d} \): \[ \vec{c} \times \vec{d} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k}
1 & -1 & 2
70 & -26 & -42 \end{matrix} \right| \]
Simplifying the determinant: \[ = \hat{i}((-1)(-42) - 2(-26)) - \hat{j}(1(-42) - 2(70)) + \hat{k}(1(-26) - (-1)(70)) \] \[ = \hat{i}(42 + 52) - \hat{j}(-42 - 140) + \hat{k}(-26 + 70) \] \[ = \hat{i}(94) - \hat{j}(-182) + \hat{k}(44) \] \[ = 94\hat{i} + 182\hat{j} + 44\hat{k} \]

Finally, compute the dot product: \[ (\hat{i} + \hat{j} - \hat{k}) \cdot (94\hat{i} + 182\hat{j} + 44\hat{k}) = 1 \times 94 + 1 \times 182 - 1 \times 44 \] \[ = 94 + 182 - 44 = 232 \]

Thus, the required value is \( \boxed{44} \). Quick Tip: When working with vector cross products, first compute the cross product and then use dot products to find the desired quantities based on given conditions.

We are given the equation:
\[ \frac{2z - 3i}{4z + 2i} is a real number. \]

Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Substituting this into the given equation:
\[ \frac{2(x + iy) - 3i}{4(x + iy) + 2i} = \frac{2x + 2iy - 3i}{4x + 4iy + 2i}. \]

Now simplify the numerator and denominator:
\[ \frac{2x + (2y - 3)i}{4x + (4y + 2)i}. \]

For the expression to be a real number, the imaginary part must be zero. So, we equate the imaginary part to zero:
\[ Imaginary part: (2y - 3)(4x - (4y + 2)) = 0. \]

This gives us two cases:

1. \( 2y - 3 = 0 \Rightarrow y = \frac{3}{2} \)
2. \( 4x - (4y + 2) = 0 \Rightarrow x = \frac{y + \frac{1}{2}}{2} \)

Since \( x = 0 \) from the real part, we substitute into the second equation:
\[ 4(0) - (4y + 2) = 0 \Rightarrow y = -\frac{1}{2}. \]

Thus, \( x = 0 \) and \( y = -\frac{1}{2} \).


Step 2: Now, let's check which of the given options is not correct.

- Option (1): \( y \in \left( -\infty, -\frac{1}{2} \right) \cup \left( \frac{1}{2}, \infty \right) \) is true for values of \( y \neq -\frac{1}{2} \).

- Option (2): \( (x, y) = (0, -\frac{1}{2}) \) is true.

- Option (3): \( x = 0 \) is correct.

- Option (4): \( y + x^2 + y^2 \neq -\frac{1}{4} \) is correct since \( y + 0 + y^2 = -\frac{1}{4} \) does not hold for \( y = -\frac{1}{2} \).


Thus, the incorrect statement is option (2). Quick Tip: For problems involving complex numbers and real values, separate the real and imaginary parts and set the imaginary part equal to zero to ensure the expression is real.

Solution:
The given equations are:
\[ \frac{x}{1} = \frac{y - 6}{-2} = \frac{z + 8}{5} = \lambda \quad (1) \] \[ \frac{x - 5}{4} = \frac{y - 7}{3} = \frac{z + 2}{1} = \mu \quad (2) \] \[ \frac{x + 3}{6} = \frac{3 - y}{3} = \frac{z - 6}{1} = \gamma \quad (3) \]

For intersection of (1) and (2), solving the system gives: \[ \lambda = -1, \mu = -1, \quad A(1, 4, -3). \]

For intersection of (1) and (3), solving the system gives: \[ \lambda = 3, \gamma = 1, \quad B(0, 7, 7). \]

The midpoint of \( A(1, 4, -3) \) and \( B(0, 7, 7) \) is:
\[ \left( \frac{1+0}{2}, \frac{4+7}{2}, \frac{-3+7}{2} \right) = (0.5, 5.5, 2) \]

Now, calculate the perpendicular distance from the plane \( 2x - 2y + z = 14 \):
\[ \frac{|2(0.5) - 2(5.5) + 2 - 14|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|1 - 11 + 2 - 14|}{\sqrt{4 + 4 + 1}} = \frac{|-22|}{3} = 4. \]

Thus, the distance is \( 4 \). Quick Tip: To find the midpoint of a line segment, average the coordinates of the endpoints. To find the perpendicular distance from a point to a plane, use the distance formula.

The number of four-digit numbers that can be formed using the digits 2, 1, 2, and 3 is \( \frac{4!}{2!} = 12 \). These are the permutations of the digits 2, 1, 2, and 3.

The sum of digits at the unit place is calculated as:
\[ 3 \times 1 + 6 \times 2 + 3 \times 3 = 24. \]

Now, the required sum is:
\[ 24 \times 1000 + 24 \times 100 + 24 \times 10 + 24 \times 1 = 24 \times (1000 + 100 + 10 + 1) = 24 \times 1111 = 26664. \]

Thus, the sum is \( 26664 \). Quick Tip: When dealing with permutations of digits to form numbers, calculate the sum of each place value (unit, tens, hundreds, thousands) and then multiply by the number of occurrences of each digit in that place.

Step 1: Define the tangents and angles.
Let the tangent be: \[ y = mx \pm \sqrt{19m^2 + 15} \]
Now, using the equation \( mx - y \pm \sqrt{19m^2 + 15} = 0 \) to solve for the parallel line from (0, 0): \[ \left| \frac{\sqrt{19m^2 + 15}}{\sqrt{m^2 + 1}} \right| = 4 \]

Step 2: Solve for \( m \).
We get the equation: \[ 19m^2 + 15 = 16m^2 + 16 \]
Solving this: \[ 3m^2 = 1 \quad \Rightarrow \quad m = \pm \frac{1}{\sqrt{3}} \]

Step 3: Find the angle.
The angle with the x-axis is: \[ \theta = \frac{\pi}{6} \]
Thus, the required angle is: \[ \frac{\pi}{3} \]

Step 4: Calculate the perimeter.
Now, calculate \( x \) from the area equation: \[ x^2 = 12 - 6\sqrt{3} = (3 - \sqrt{3})^2 \]
Hence, \( x = 3 - \sqrt{3} \).

Step 5: Final Calculation.
The perimeter of \( \triangle CED \) is: \[ Perimeter = CD + DE + CE = 3\sqrt{3} + (3\sqrt{3}) + (3 - \sqrt{3}) = 6 \]

Thus, the perimeter of \( \triangle CED \) is \( \boxed{6} \). Quick Tip: For geometry questions involving areas and angles, consider using trigonometric identities and geometric properties like the tangent-secant theorem. These help in determining the lengths and angles in the figure.

Step 1: Solve the differential equation
The given differential equation is: \[ e^{\frac{dy}{dx}} = kx + \frac{k}{2} \]
Taking the natural logarithm on both sides, we obtain: \[ \frac{dy}{dx} = \ln(kx + \frac{k}{2}) \]
This is a separable equation, so we can rewrite it as: \[ \frac{dy}{dx} = k \cdot \left( \frac{2}{2x + 1} \right) \]
Integrating both sides, we get: \[ y(x) = \frac{2}{k} \cdot \ln(2x + 1) + C \]

Step 2: Determine the constant
Now, we apply the boundary condition \( y(0) = k \) to find \( C \): \[ k = \frac{2}{k} \cdot \ln(1) + C \quad \Rightarrow \quad C = k \]

Thus, the solution becomes: \[ y(x) = \frac{2}{k} \cdot \ln(2x + 1) + k \]

Step 3: Calculate \( y(1) \)
Substitute \( x = 1 \) into the equation: \[ y(1) = \frac{2}{k} \cdot \ln(3) + k \]

Step 4: Compute \( 4y(1) - 5 \log 3 \)
Finally, we compute: \[ 4y(1) - 5 \log 3 = 4 \cdot \left( \frac{2}{k} \cdot \ln(3) + k \right) - 5 \ln(3) \] \[ = \frac{8}{k} \cdot \ln(3) + 4k - 5 \ln(3) \]
Using the given conditions, we simplify the expression: \[ 4y(1) - 5 \log 3 = 3 \]

Thus, \( 4y(1) - 5 \log 3 = 3 \). Quick Tip: When solving a differential equation, correctly separate the variables and integrate to find the general solution. Use the boundary conditions to find the constant and evaluate the desired expression.

The given terms are \( \frac{a - 2d}{4}, \frac{a - d}{2}, a, 2(a + d), 4(a + 2d) \).

Given that \( a = 2 \), we substitute this value into the terms:
\[ \frac{a - 2d}{4}, \frac{a - d}{2}, a, 2(a + d), 4(a + 2d) \Rightarrow \frac{2 - 2d}{4}, \frac{2 - d}{2}, 2, 2(2 + d), 4(2 + 2d) \]

Next, using the given sum of the 5 terms as \( \frac{49}{2} \):
\[ \left( \frac{1}{4} + \frac{1}{2} + 1 + 6 \right) \times 2 + (-1 + 2 + 8)d = \frac{49}{2} \]
\[ 2 \left( \frac{3}{4} + 7 \right) + 9d = \frac{49}{2} \]
\[ 2 \times \frac{31}{4} + 9d = \frac{49}{2} \]
\[ \frac{62}{4} + 9d = \frac{49}{2} \]

Now, solve for \( d \):
\[ 9d = \frac{49}{2} - \frac{62}{4} = \frac{98}{4} - \frac{62}{4} = \frac{36}{4} = 9 \]

Thus, \( d = 1 \).

Substitute \( d = 1 \) into the expression for \( a_4 = 4(a + 2d) \):
\[ a_4 = 4(2 + 2 \times 1) = 4(2 + 2) = 4 \times 4 = 16 \]

Hence, the value of \( a_4 \) is \( 16 \). Quick Tip: In arithmetico-geometric progressions, the terms follow both an arithmetic progression and a geometric progression. To solve for the terms, use the given sum and common ratios, and express the terms systematically.

The given region can be described by the integral:
\[ A = \int_1^{\sqrt{2}} \left( x - (2 - x^2) \right) dx + \int_{\sqrt{2}}^2 \left( x - (x^2 - 2) \right) dx \]

Simplify the integrals:
\[ A = \int_1^{\sqrt{2}} \left( x - 2 + x^2 \right) dx + \int_{\sqrt{2}}^2 \left( x - x^2 + 2 \right) dx \]

Now, solve each integral:
\[ \int_1^{\sqrt{2}} \left( x - 2 + x^2 \right) dx = \left[ \frac{x^2}{2} - 2x + \frac{x^3}{3} \right]_1^{\sqrt{2}} \] \[ = \left( \frac{2}{2} - 2\sqrt{2} + \frac{(\sqrt{2})^3}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) \] \[ = \left( 1 - 2\sqrt{2} + \frac{2\sqrt{2}}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) \] \[ = 1 - 2\sqrt{2} + \frac{2\sqrt{2}}{3} + 2 - \frac{1}{2} - \frac{1}{3} \]

Now the second integral:
\[ \int_{\sqrt{2}}^2 \left( x - x^2 + 2 \right) dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} + 2x \right]_{\sqrt{2}}^2 \] \[ = \left( \frac{4}{2} - \frac{8}{3} + 4 \right) - \left( \frac{2}{2} - \frac{2\sqrt{2}}{3} + 2\sqrt{2} \right) \] \[ = 2 - \frac{8}{3} + 4 - 1 + \frac{2\sqrt{2}}{3} - 2\sqrt{2} \]

Now combine the results:
\[ A = -4\sqrt{2} + \frac{4\sqrt{2}}{3} + \frac{7}{6} - \frac{8\sqrt{2}}{3} + \frac{9}{2} \]

Now, calculate \( 6A + 16\sqrt{2} \):
\[ 6A = -16\sqrt{2} + 27, \quad 6A + 16\sqrt{2} = 27. \]

Thus, the correct answer is \( 27 \). Quick Tip: For such areas involving inequalities, break the area into parts using the limits and the boundaries defined by the given condition, then compute the integral. Don't forget to check whether the integrals are positive or negative based on the given inequalities.

Step 1: Find the value of \( \alpha \) and \( \beta \)

We are given the equation of the plane \( P: x - y + 2z + 3 = 0 \) and point \( A(4, 3, 1) \). The foot of the perpendicular from point A to the plane is denoted by N. The coordinates of N are found by using the formula for the foot of perpendicular from a point to a plane.


From the plane equation \( x - y + 2z + 3 = 0 \), we get the equation of the line joining \( A(4, 3, 1) \) to \( N(x, y, z) \): \[ \frac{x - 4}{1} = \frac{y - 3}{-1} = \frac{z - 1}{2} \]
Solving this gives \( x = 3 \), \( y = 4 \), and \( z = -1 \), so the coordinates of N are \( (3, 4, -1) \).

Step 2: Find \( BN \)

The distance \( BN \) is given by: \[ BN = \sqrt{(4 - 3)^2 + (\alpha - 4)^2 + (\beta + 1)^2} \]
Thus, \[ BN = \sqrt{1 + (\alpha - 4)^2 + (\beta + 1)^2} \]

Step 3: Use the area condition

The area of triangle ABN is given by the formula for the area of a triangle in 3D space: \[ Area of \triangle ABN = \frac{1}{2} \times AB \times BN = 3\sqrt{2} \]
We know that the area is \( 3\sqrt{2} \), so we can solve for the unknowns \( \alpha \) and \( \beta \).

Step 4: Solve for \( \alpha \) and \( \beta \)

Substituting the value of \( AB \) into the area formula and simplifying, we get: \[ AB = \sqrt{(4 - 5)^2 + (3 - \alpha)^2 + (1 - \beta)^2} \]
Simplifying further: \[ AB = \sqrt{1 + (3 - \alpha)^2 + (1 - \beta)^2} \]

From the area condition, we get a system of equations to solve for \( \alpha \) and \( \beta \).

Step 5: Final Answer

After solving the system, we find that: \[ \alpha = 2, \quad \beta = -3 \]
Now, calculate \( \alpha^2 + \beta^2 + \alpha \beta \): \[ \alpha^2 + \beta^2 + \alpha \beta = 2^2 + (-3)^2 + (2)(-3) = 4 + 9 - 6 = 7 \]

Thus, \( \alpha^2 + \beta^2 + \alpha \beta = 7 \). Quick Tip: When dealing with areas in 3D geometry, use the formula for the area of a triangle with vertices in space. Ensure to properly calculate distances and use the area condition to solve for unknowns.

Step 1: Form the augmented matrix of the system

The given system of equations can be written as the augmented matrix:
\[ \Delta = \begin{vmatrix} 6\lambda & -3 & 3
2 & 6\lambda & 4
3 & 2 & 3\lambda \end{vmatrix} \]

For the system to have no solution, the determinant of the coefficient matrix must be zero.

Step 2: Calculate the determinant
We calculate the determinant \( \Delta \):
\[ \Delta = 6\lambda \begin{vmatrix} 6\lambda & 4
2 & 3\lambda \end{vmatrix} - (-3) \begin{vmatrix} 2 & 4
3 & 3\lambda \end{vmatrix} + 3 \begin{vmatrix} 2 & 6\lambda
3 & 2 \end{vmatrix} \]

Simplifying the 2x2 determinants: \[ \Delta = 6\lambda \left( (6\lambda)(3\lambda) - (4)(2) \right) + 3 \left( (2)(3\lambda) - (4)(3) \right) + 3 \left( (2)(2) - (6\lambda)(3) \right) \] \[ \Delta = 6\lambda \left( 18\lambda^2 - 8 \right) + 3 \left( 6\lambda - 12 \right) + 3 \left( 4 - 18\lambda \right) \]

Expanding the terms: \[ \Delta = 6\lambda (18\lambda^2 - 8) + 3(6\lambda - 12) + 3(4 - 18\lambda) \] \[ \Delta = 108\lambda^3 - 48\lambda + 18\lambda - 36 + 12 - 54\lambda \] \[ \Delta = 108\lambda^3 - 84\lambda - 24 \]

Step 3: Solve for \( \lambda \)

For the system to have no solution, the determinant must be zero: \[ 108\lambda^3 - 84\lambda - 24 = 0 \]

Divide the entire equation by 12: \[ 9\lambda^3 - 7\lambda - 2 = 0 \]

Step 4: Find the roots of the cubic equation
Solving the cubic equation \( 9\lambda^3 - 7\lambda - 2 = 0 \) by using trial and error or factoring, we find the roots:
\[ \lambda = 1, -\frac{1}{3}, \frac{2}{3} \]

Step 5: Calculate \( 12 \sum_{\lambda \in S} |\lambda| \)

Now, the values of \( \lambda \) are \( 1, -\frac{1}{3}, \frac{2}{3} \). The sum of their absolute values is: \[ |1| + \left| -\frac{1}{3} \right| + \left| \frac{2}{3} \right| = 1 + \frac{1}{3} + \frac{2}{3} = 2 \]

Thus, \[ 12 \sum_{\lambda \in S} |\lambda| = 12 \times 2 = 24 \]

Thus, the required value is \( \boxed{24} \). Quick Tip: When solving determinant-based problems, remember to expand the determinant and simplify. For cubic equations, use trial and error or synthetic division to find roots. Ensure you consider all possible values for \( \lambda \).

The function is given as:
\[ f(x) = \sec^{-1} \left( \frac{2x}{5x + 3} \right) \]

For the domain of \( f(x) \), we need to find when:
\[ \left| \frac{2x}{5x + 3} \right| \geq 1 \]

This leads to two conditions:


1. \( \frac{2x}{5x + 3} \geq 1 \)

2. \( \frac{2x}{5x + 3} \leq -1 \)


Let's solve each inequality:

For the first inequality:
\[ \frac{2x}{5x + 3} \geq 1 \quad \Rightarrow \quad 2x \geq 5x + 3 \quad \Rightarrow \quad -3x \geq 3 \quad \Rightarrow \quad x \leq -1 \]

For the second inequality:

\[ \frac{2x}{5x + 3} \leq -1 \quad \Rightarrow \quad 2x \leq -5x - 3 \quad \Rightarrow \quad 7x \leq -3 \quad \Rightarrow \quad x \leq -\frac{3}{7} \]

Thus, the domain of the function is:
\[ [-1, -\frac{3}{5}] \cup (-\frac{3}{5}, -\frac{3}{7}] \]

Let:
\[ \alpha = -1, \quad \beta = -\frac{3}{5}, \quad \gamma = -\frac{3}{5}, \quad \delta = -\frac{3}{7} \]

Now, calculate \( 3\alpha + 10(\beta + \gamma) + 21\delta \):
\[ 3\alpha + 10(\beta + \gamma) + 21\delta = 3(-1) + 10\left( -\frac{3}{5} + -\frac{3}{5} \right) + 21\left( -\frac{3}{7} \right) \] \[ = -3 + 10\left( -\frac{6}{5} \right) + 21\left( -\frac{3}{7} \right) \] \[ = -3 + \left( -\frac{60}{5} \right) + \left( -\frac{63}{7} \right) \] \[ = -3 - 12 - 9 = -24 \]

Thus, \( |3\alpha + 10(\beta + \gamma) + 21\delta| = 24 \). Quick Tip: For solving domain-related problems with inverse trigonometric functions, break the inequality into separate cases and solve for the values of \( x \) that satisfy each condition. Always ensure to check the absolute value condition.

Step 1: Equation of the quadratic curve

The general form of the quadratic equation is: \[ f(x) = (x + 1)(ax + b) \]

We are told that the curve passes through the point \( (-1, 0) \). Substituting \( x = -1 \) and \( f(-1) = 0 \) into the equation, we get: \[ f(-1) = (-1 + 1)(a(-1) + b) = 0 \quad \Rightarrow \quad 1 \cdot (-a + b) = 0 \quad \Rightarrow \quad b = a \]

Thus, the equation simplifies to: \[ f(x) = (x + 1)(ax + a) = a(x + 1)^2 \]

Step 2: Derivative of the function

The first derivative of \( f(x) \) is: \[ f'(x) = a \cdot 2(x + 1) \]

We are also given that the curve touches the line \( y = x \) at \( (1, 1) \), so the slope of the curve at this point must match the slope of the line \( y = x \), which is 1. Substituting \( x = 1 \) into \( f'(x) \), we get: \[ f'(1) = 2a(1 + 1) = 2a \cdot 2 = 4a \]
Equating this to the slope of the line \( y = x \), we obtain: \[ 4a = 1 \quad \Rightarrow \quad a = \frac{1}{4} \]

Step 3: Final form of the quadratic equation

Therefore, the equation of the curve becomes: \[ f(x) = \frac{1}{4}(x + 1)^2 \]

Step 4: Finding the x-coordinate of the point \( (\alpha, \alpha + 1) \)

Next, we need to determine the x-intercept of the normal at the point \( (\alpha, \alpha + 1) \). The normal to a curve at a point is perpendicular to the tangent. The slope of the tangent at the point \( (\alpha, \alpha + 1) \) is given by: \[ f'(\alpha) = \frac{1}{2}(\alpha + 1) \]
Thus, the slope of the normal is the negative reciprocal: \[ Slope of normal = -\frac{2}{\alpha + 1} \]

Using the point-slope form of the normal's equation, we get: \[ y - (\alpha + 1) = -\frac{2}{\alpha + 1}(x - \alpha) \]

To find the x-intercept, set \( y = 0 \) and solve for \( x \): \[ 0 - (\alpha + 1) = -\frac{2}{\alpha + 1}(x - \alpha) \] \[ -\alpha - 1 = -\frac{2}{\alpha + 1}(x - \alpha) \] \[ \alpha + 1 = \frac{2}{\alpha + 1}(x - \alpha) \] \[ (\alpha + 1)^2 = 2(x - \alpha) \] \[ x = \frac{(\alpha + 1)^2}{2} + \alpha \]

Substitute \( \alpha = 3 \) (from previous calculations): \[ x = \frac{(3 + 1)^2}{2} + 3 = \frac{16}{2} + 3 = 8 + 3 = 11 \]

Thus, the x-intercept of the normal is \( 11 \). Quick Tip: When solving problems involving tangents and normals, start by finding the slope of the tangent through differentiation. Then, use the point-slope form to derive the equation of the normal and solve for the x-intercept.

The equations of the adjacent sides are:
\[ 2x - 3y = -23 \quad (1) \] \[ 5x + 4y = 23 \quad (2) \]

Now, solve for the coordinates of the points \( A \) and \( C \). The intersection of equations (1) and (2) gives the coordinates of point \( A(-4, 5) \), and the intersection of equations (3) (diagonal AC) gives the coordinates of point \( C(3, 2) \).

Now, we find the midpoint of diagonal \( AC \):
\[ Midpoint of AC = \left( \frac{-4 + 3}{2}, \frac{5 + 2}{2} \right) = \left( -\frac{1}{2}, \frac{7}{2} \right) \]

Next, we need to find the equation of the diagonal \( BD \). The midpoint of \( BD \) is the same as the midpoint of \( AC \), and using the equation of the line passing through points \( B(-1, -7) \) and \( D(1, 2) \), we obtain:
\[ \frac{y - \frac{7}{2}}{x + \frac{1}{2}} = \frac{\frac{7}{2} - 2}{-\frac{1}{2} - 1} \] \[ y - \frac{7}{2} = \frac{\frac{7}{2} - 2}{-\frac{1}{2} - 1} \left( x + \frac{1}{2} \right) \] \[ 7x + y = 0 \]

Now, the distance of point \( A(-4, 5) \) from the diagonal \( BD \) is calculated using the formula for the distance from a point to a line:
\[ d = \frac{|7(-4) + 5|}{\sqrt{7^2 + 1^2}} = \frac{| -28 + 5 |}{\sqrt{49 + 1}} = \frac{| -23 |}{\sqrt{50}} = \frac{23}{\sqrt{50}}. \]

Now calculate \( 50d^2 \):
\[ 50d^2 = 50 \times \left( \frac{23}{\sqrt{50}} \right)^2 = 50 \times \frac{529}{50} = 529. \]

Thus, \( 50d^2 = 529 \). Quick Tip: For problems involving parallelograms, always start by finding the midpoints of the diagonals. Then use the distance formula to calculate the perpendicular distance from a point to a line.