GUJCET 2026 Mathematics Question Paper with Answer and Solution

(C) The relation is defined as $R = \{(a, b) : a = b - 2, b > 6\}$.
Given $b > 6$,let us test the options.
For option $(A)$,$(8, 6)$: Here $b = 6$,which does not satisfy $b > 6$.
For option $(B)$,$(3, 8)$: Here $a = 3$ and $b = 8$. Checking the condition $a = b - 2$,we get $3 = 8 - 2 = 6$,which is false.
For option $(C)$,$(6, 8)$: Here $a = 6$ and $b = 8$. Checking the condition $a = b - 2$,we get $6 = 8 - 2 = 6$,which is true. Also,$b = 8 > 6$ is satisfied.
For option $(D)$,$(8, 7)$: Here $a = 8$ and $b = 7$. Checking the condition $a = b - 2$,we get $8 = 7 - 2 = 5$,which is false.
Therefore,$(6, 8) \in R$ is the correct answer.

(D) Given $e^y(x+1) = 1$.
Taking the natural logarithm on both sides,we get: $y + \ln(x+1) = 0 \implies y = -\ln(x+1)$.
Differentiating with respect to $x$: $\frac{dy}{dx} = -\frac{1}{x+1} = -(x+1)^{-1}$.
Differentiating again with respect to $x$: $\frac{d^2y}{dx^2} = -(-1)(x+1)^{-2} = \frac{1}{(x+1)^2} = \left(\frac{1}{x+1}\right)^2$.
Now,calculating the expression: $\frac{d^2y}{dx^2} - \left(\frac{dy}{dx}\right)^2 = \left(\frac{1}{x+1}\right)^2 - \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2} - \frac{1}{(x+1)^2} = 0$.

(A) Let $I = \int_{\pi/6}^{\pi/3} \frac{1}{1+\sqrt{\cot x}} dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$.
Using the definite integral property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$.
Thus,$I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin(\pi/2-x)}}{\sqrt{\sin(\pi/2-x)} + \sqrt{\cos(\pi/2-x)}} dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$.
Adding the two expressions for $I$:
$2I = \int_{\pi/6}^{\pi/3} \left( \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) dx = \int_{\pi/6}^{\pi/3} 1 dx$.
$2I = [x]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$I = \frac{\pi}{12}$.

(B) We know that the trigonometric identity is $\cos^2 \theta - \sin^2 \theta = \cos 2\theta$.
Therefore,$\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}) = -\cos(2 \cdot \frac{x}{2}) = -\cos x$.
Now,the integral becomes $I = \int_0^{\pi} -\cos x \, dx$.
Integrating $-\cos x$ with respect to $x$,we get $-\sin x$.
Applying the limits from $0$ to $\pi$,we have $I = [-\sin x]_0^{\pi}$.
$I = -(\sin \pi - \sin 0)$.
Since $\sin \pi = 0$ and $\sin 0 = 0$,we get $I = -(0 - 0) = 0$.

(D) To evaluate the integral $I = \int \frac{dx}{\sqrt{9-8x-4x^2}}$,we first complete the square for the quadratic expression in the denominator.
$9 - 8x - 4x^2 = 9 - (4x^2 + 8x) = 9 - 4(x^2 + 2x)$.
Adding and subtracting $1$ inside the bracket: $9 - 4(x^2 + 2x + 1 - 1) = 9 - 4(x+1)^2 + 4 = 13 - (2x+2)^2$.
Thus,the integral becomes $I = \int \frac{dx}{\sqrt{(\sqrt{13})^2 - (2x+2)^2}}$.
Using the standard formula $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + C$,where $u = 2x+2$ and $du = 2dx$ (so $dx = \frac{du}{2}$):
$I = \frac{1}{2} \int \frac{du}{\sqrt{(\sqrt{13})^2 - u^2}} = \frac{1}{2} \sin^{-1}(\frac{u}{\sqrt{13}}) + C$.
Substituting $u = 2x+2$,we get $I = \frac{1}{2} \sin^{-1}(\frac{2x+2}{\sqrt{13}}) + C$.

(A) We have the integral $I = \int \sec^2 x \csc^2 x \, dx$.
Using the identities $\sec^2 x = \frac{1}{\cos^2 x}$ and $\csc^2 x = \frac{1}{\sin^2 x}$,we get:
$I = \int \frac{1}{\cos^2 x \sin^2 x} \, dx$.
Since $1 = \sin^2 x + \cos^2 x$,we can write:
$I = \int \frac{\sin^2 x + \cos^2 x}{\cos^2 x \sin^2 x} \, dx$.
$I = \int \left( \frac{\sin^2 x}{\cos^2 x \sin^2 x} + \frac{\cos^2 x}{\cos^2 x \sin^2 x} \right) \, dx$.
$I = \int (\sec^2 x + \csc^2 x) \, dx$.
Integrating term by term,we get:
$I = \tan x - \cot x + C$.

(A) To find the interval where the function $y = x^2 e^{-x}$ is decreasing,we calculate its derivative $\frac{dy}{dx}$.
Using the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) = 2x e^{-x} - x^2 e^{-x} = x e^{-x} (2 - x)$.
$A$ function is decreasing when $\frac{dy}{dx} < 0$.
Since $e^{-x} > 0$ for all real $x$,the inequality $\frac{dy}{dx} < 0$ simplifies to $x(2 - x) < 0$.
Multiplying by $-1$ reverses the inequality: $x(x - 2) > 0$.
This inequality holds when $x < 0$ or $x > 2$.
Thus,the function is decreasing on the interval $(-\infty, 0) \cup (2, \infty)$.

(C) We know that the absolute value function $|x+1|$ is always non-negative,meaning $|x+1| \ge 0$ for all $x \in R$.
Multiplying by $-1$ reverses the inequality: $-|x+1| \le 0$.
Adding $3$ to both sides gives: $-|x+1| + 3 \le 0 + 3$,which simplifies to $f(x) \le 3$.
The maximum value is attained when the term $-|x+1|$ is equal to $0$,which happens at $x = -1$.
Therefore,the maximum value of the function is $3$.

(D) Marginal Revenue $(MR)$ is defined as the derivative of the total revenue function $R(x)$ with respect to $x$.
$MR = \frac{dR}{dx} = \frac{d}{dx}(3x^2 + 36x + 5)$
Applying the power rule of differentiation:
$MR = 6x + 36$
Now,substitute $x = 15$ into the expression for $MR$:
$MR = 6(15) + 36$
$MR = 90 + 36 = 126$
Therefore,the marginal revenue when $x = 15$ is $126$.

(B) Let $C = AB - BA$.
Taking the transpose of $C$,we get $C^T = (AB - BA)^T$.
Using the property $(X - Y)^T = X^T - Y^T$ and $(XY)^T = Y^T X^T$,we have $C^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T$.
Since $A$ and $B$ are skew-symmetric,$A^T = -A$ and $B^T = -B$.
Substituting these values,$C^T = (-B)(-A) - (-A)(-B) = BA - AB$.
Factoring out a negative sign,$C^T = -(AB - BA) = -C$.
Since $C^T = -C$,the matrix $AB - BA$ is a skew-symmetric matrix.

(C) Given $A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$.
We calculate $A^2 = A \times A = \begin{bmatrix} a & b \\ c & -a \end{bmatrix} \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$.
Performing matrix multiplication: $A^2 = \begin{bmatrix} a^2 + bc & ab - ab \\ ac - ac & bc + a^2 \end{bmatrix} = \begin{bmatrix} a^2 + bc & 0 \\ 0 & a^2 + bc \end{bmatrix}$.
Given that $A^2 = I$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,we equate the matrices:
$\begin{bmatrix} a^2 + bc & 0 \\ 0 & a^2 + bc \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Comparing the elements,we get $a^2 + bc = 1$.
Rearranging the equation,we get $1 - a^2 - bc = 0$.

(B) We know that $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$.
Substituting this into the expression: $2 \sin^{-1}(\frac{1}{2}) = 2(\frac{\pi}{6}) = \frac{\pi}{3}$.
Now,the expression becomes $\tan^{-1} [2 \cos(\frac{\pi}{3})]$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have $\tan^{-1} [2 \times \frac{1}{2}] = \tan^{-1}(1)$.
Since $\tan(\frac{\pi}{4}) = 1$,the final value is $\frac{\pi}{4}$.

(A) The principal value branch of $\sin^{-1} x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
Since $\frac{3\pi}{5} > \frac{\pi}{2}$,we must rewrite $\sin(\frac{3\pi}{5})$ using the identity $\sin(\pi - \theta) = \sin(\theta)$.
Thus,$\sin(\frac{3\pi}{5}) = \sin(\pi - \frac{3\pi}{5}) = \sin(\frac{2\pi}{5})$.
Since $\frac{2\pi}{5}$ lies in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$,we have $\sin^{-1}(\sin \frac{2\pi}{5}) = \frac{2\pi}{5}$.

(B) The principal value branch of the inverse cosine function $\cos^{-1}(x)$ is defined as $[0, \pi]$.
Therefore,if $y = \cos^{-1}(x)$,then $y$ must lie in the interval $[0, \pi]$.

(C) Testing for one-one: $f(1) = \frac{1+1}{2} = 1$ and $f(2) = \frac{2}{2} = 1$. Since $f(1) = f(2)$ for $1 \neq 2$,the function is not one-one (it is many-one).
Testing for onto: For any $y \in N$,if $y$ is odd,we can have $n = 2y-1$,then $f(2y-1) = \frac{(2y-1)+1}{2} = y$. If $y$ is even,we can have $n = 2y$,then $f(2y) = \frac{2y}{2} = y$. Thus,for every element in the codomain,there is a pre-image in the domain. Therefore,the function is onto.

(A) Given the equation $P(A) + P(B) - P(A \cap B) = P(A)$.
Subtracting $P(A)$ from both sides,we get $P(B) - P(A \cap B) = 0$,which implies $P(B) = P(A \cap B)$.
This equality indicates that $B \subseteq A$,meaning all outcomes of event $B$ are contained within event $A$.
By the definition of conditional probability,$P(A|B) = \frac{P(A \cap B)}{P(B)}$.
Since $P(A \cap B) = P(B)$,we substitute this into the formula:
$P(A|B) = \frac{P(B)}{P(B)} = 1$ (assuming $P(B) \neq 0$).

(C) The total number of cards is $52$. There are $4$ kings and $4$ aces in a deck.
Step $1$: Probability of drawing the first king = $4/52 = 1/13$.
Step $2$: Probability of drawing the second king (without replacement) = $3/51 = 1/17$.
Step $3$: Probability of drawing the third card as an ace (without replacement) = $4/50 = 2/25$.
Total probability = $(4/52) \times (3/51) \times (4/50) = (1/13) \times (1/17) \times (2/25) = 2 / 5525$.

(C) Let $u = \log_{2025} x = \frac{\log x}{\log 2025}$.
Then $y = \log_{2026} u = \frac{\log u}{\log 2026}$.
Applying the chain rule:
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u \log 2026} \cdot \frac{d}{dx}\left(\frac{\log x}{\log 2025}\right)$.
$\frac{dy}{dx} = \frac{1}{u \log 2026} \cdot \frac{1}{x \log 2025}$.
Substituting $u = \frac{\log x}{\log 2025}$:
$\frac{dy}{dx} = \frac{1}{(\frac{\log x}{\log 2025}) \log 2026 \cdot x \log 2025} = \frac{1}{x \log x \log 2026}$.

(A) Given $x = at^2$ and $y = 2at$.
First,find $\frac{dx}{dt} = 2at$ and $\frac{dy}{dt} = 2a$.
Then,$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a}{2at} = \frac{1}{t}$.
Now,differentiate with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{1}{t}) = \frac{d}{dt}(\frac{1}{t}) \cdot \frac{dt}{dx} = (-\frac{1}{t^2}) \cdot \frac{1}{2at} = -\frac{1}{2at^3}$.
We know $x = at^2$,so $t^2 = \frac{x}{a}$. Also $y = 2at$,so $t = \frac{y}{2a}$.
Substitute $t^2 = \frac{x}{a}$ and $t = \frac{y}{2a}$ into the expression:
$\frac{d^2y}{dx^2} = -\frac{1}{2a(t^2)(t)} = -\frac{1}{2a(\frac{x}{a})(\frac{y}{2a})} = -\frac{1}{\frac{xy}{a}} = -\frac{a}{xy}$.

(C) For a function $f(x)$ to be continuous at $x = a$,the left-hand limit $(LHL)$,right-hand limit $(RHL)$,and the value of the function at $x = a$ must be equal.
Here,$a = \pi$.
$LHL$: $\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} (kx + 1) = k\pi + 1$.
$RHL$: $\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (\cos x) = \cos(\pi) = -1$.
Since the function is continuous at $x = \pi$,we have $LHL$ = $RHL$.
Therefore,$k\pi + 1 = -1$.
$k\pi = -2$.
$k = -\frac{2}{\pi}$.

(B) The inverse of a matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is given by $A^{-1} = \frac{1}{|A|} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Given $A = \begin{bmatrix} 2 & 3 \\ 1 & -4 \end{bmatrix}$,the determinant $|A| = (2)(-4) - (3)(1) = -8 - 3 = -11$.
Thus,$A^{-1} = \frac{1}{-11} \begin{bmatrix} -4 & -3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4/11 & 3/11 \\ 1/11 & -2/11 \end{bmatrix}$.
Comparing this with the given $A^{-1} = \begin{bmatrix} a & 3/11 \\ 1/11 & b \end{bmatrix}$,we get $a = 4/11$ and $b = -2/11$.
Therefore,$a+b = 4/11 + (-2/11) = 2/11$.

(A) Given $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$,then $A^2 = \begin{bmatrix} 2^2 & 0 & 0 \\ 0 & 3^2 & 0 \\ 0 & 0 & 4^2 \end{bmatrix} = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 16 \end{bmatrix}$.
Given $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -3 \end{bmatrix}$,then $B^2 = \begin{bmatrix} 1^2 & 0 & 0 \\ 0 & (-2)^2 & 0 \\ 0 & 0 & (-3)^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 9 \end{bmatrix}$.
Therefore,$A^2 + B^2 = \begin{bmatrix} 4+1 & 0 & 0 \\ 0 & 9+4 & 0 \\ 0 & 0 & 16+9 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 0 \\ 0 & 13 & 0 \\ 0 & 0 & 25 \end{bmatrix}$.

(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula: $\text{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = 35$.
Substituting the given vertices $(2, -6)$,$(5, 4)$,and $(k, 4)$ into the formula:
$\frac{1}{2} |2(4-4) + 5(4 - (-6)) + k(-6-4)| = 35$
$\frac{1}{2} |2(0) + 5(10) + k(-10)| = 35$
$|50 - 10k| = 70$
This gives two possible cases:
Case $1$: $50 - 10k = 70 \implies -10k = 20 \implies k = -2$.
Case $2$: $50 - 10k = -70 \implies -10k = -120 \implies k = 12$.
Thus,the possible values for $k$ are $12$ and $-2$.

(A) Given $A = \begin{bmatrix} 3 & -2 \\ 4 & 2 \end{bmatrix}$.
First,calculate the trace of $A$: $\text{tr}(A) = 3 + 2 = 5$.
Next,calculate the determinant of $A$: $|A| = (3)(2) - (-2)(4) = 6 + 8 = 14$.
According to the Cayley-Hamilton theorem,every square matrix satisfies its own characteristic equation: $A^2 - \text{tr}(A)A + |A|I = 0$.
Substituting the values,we get $A^2 - 5A + 14I = 0$.
Rearranging the equation to solve for $A^2$,we get $A^2 = 5A - 14I$.

(B) The given integral is $I = \int e^x \left( \frac{1-x}{1+x^2} \right)^2 dx$.
Expanding the numerator,we get $I = \int e^x \frac{1-2x+x^2}{(1+x^2)^2} dx$.
This can be rewritten as $I = \int e^x \left[ \frac{1+x^2}{(1+x^2)^2} - \frac{2x}{(1+x^2)^2} \right] dx$.
$I = \int e^x \left[ \frac{1}{1+x^2} - \frac{2x}{(1+x^2)^2} \right] dx$.
We know the standard integral formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.
Here,let $f(x) = \frac{1}{1+x^2}$.
Then,$f'(x) = \frac{d}{dx} (1+x^2)^{-1} = -1(1+x^2)^{-2} \cdot (2x) = -\frac{2x}{(1+x^2)^2}$.
Since the integrand is in the form $e^x [f(x) + f'(x)]$,the solution is $e^x f(x) + C = \frac{e^x}{1+x^2} + C$.

(C) Given integral is $I = \int \frac{e^{2025+x} - e^{2025-x}}{e^{2026+x} + e^{2026-x}} dx$.
Factor out $e^{2025}$ from the numerator and $e^{2026}$ from the denominator:
$I = \int \frac{e^{2025}(e^x - e^{-x})}{e^{2026}(e^x + e^{-x})} dx = \frac{1}{e} \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx$.
Let $u = e^x + e^{-x}$.
Then,the derivative is $du = (e^x - e^{-x}) dx$.
Substituting these into the integral:
$I = \frac{1}{e} \int \frac{1}{u} du = \frac{1}{e} \ln |u| + C$.
Substituting back $u = e^x + e^{-x}$,we get:
$I = \frac{1}{e} \ln |e^x + e^{-x}| + C$.

(D) The given equation of the ellipse is $4x^2 + 9y^2 = 144$.
Dividing both sides by $144$,we get $\frac{4x^2}{144} + \frac{9y^2}{144} = 1$,which simplifies to $\frac{x^2}{36} + \frac{y^2}{16} = 1$.
This is the standard form of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,where $a^2 = 36$ (so $a = 6$) and $b^2 = 16$ (so $b = 4$).
The total area of an ellipse is given by the formula $A = \pi ab$.
Substituting the values,the total area is $A = \pi \times 6 \times 4 = 24\pi$.
Since the ellipse is symmetric about both axes,the area in the first quadrant is one-fourth of the total area.
Therefore,the required area = $\frac{1}{4} \times 24\pi = 6\pi$.

(C) The function $y = x|x|$ is defined as $y = x^2$ for $x \ge 0$ and $y = -x^2$ for $x < 0$.
Since we are calculating the area,we take the absolute value of the function: $\int_{-1}^1 |x|x|| dx$.
This can be split into two intervals: $\int_{-1}^0 |-x^2| dx + \int_{0}^1 |x^2| dx$.
Since $|-x^2| = x^2$ and $|x^2| = x^2$,the integral becomes $\int_{-1}^0 x^2 dx + \int_{0}^1 x^2 dx$.
Evaluating the integrals: $[\frac{x^3}{3}]_{-1}^0 + [\frac{x^3}{3}]_{0}^1$.
$= (0 - (-1/3)) + (1/3 - 0) = 1/3 + 1/3 = 2/3$.

(D) To find the order and degree,we must first eliminate the radicals by raising both sides to the power of $6$ (the least common multiple of $2$ and $3$).
Given equation: $(1 + (y'')^2)^{1/2} = (x + (y')^3)^{1/3}$.
Raising both sides to the power of $6$:
$((1 + (y'')^2)^{1/2})^6 = ((x + (y')^3)^{1/3})^6$
$(1 + (y'')^2)^3 = (x + (y')^3)^2$.
Expanding the left side: $1 + 3(y'')^2 + 3(y'')^4 + (y'')^6 = (x + (y')^3)^2$.
The highest order derivative present is $y'' = \frac{d^2y}{dx^2}$,so the order is $2$.
The exponent of the highest order derivative after rationalizing the equation is $6$. Therefore,the degree is $6$.

(D) The general solution of a differential equation of order $n$ contains $n$ arbitrary constants.
$A$ particular solution is obtained by assigning specific values to these arbitrary constants.
Therefore,a particular solution contains $0$ arbitrary constants.

(B) The given differential equation is $\frac{dy}{dx} = e^{x+y} = e^x \cdot e^y$.
By separating the variables,we get $e^{-y} \, dy = e^x \, dx$.
Integrating both sides,we have $\int e^{-y} \, dy = \int e^x \, dx$.
This results in $-e^{-y} = e^x + C'$,where $C'$ is the constant of integration.
Rearranging the terms,we get $e^x + e^{-y} = -C'$.
Letting $C = -C'$,the general solution is $e^x + e^{-y} = C$.

(B) The magnitude of the difference of two vectors is given by the formula: $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.
Substituting the given values $|\vec{a}| = 2$,$|\vec{b}| = 3$,and $\vec{a} \cdot \vec{b} = 4$ into the formula:
$|\vec{a} - \vec{b}|^2 = (2)^2 + (3)^2 - 2(4)$
$|\vec{a} - \vec{b}|^2 = 4 + 9 - 8$
$|\vec{a} - \vec{b}|^2 = 5$
Taking the square root on both sides,we get $|\vec{a} - \vec{b}| = \sqrt{5}$.

(D) The area of a triangle with vertices $A$,$B$,and $C$ is given by the formula: $\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}|$.
First,we find the vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$.
$\vec{AC} = (1-1)\hat{i} + (5-1)\hat{j} + (5-2)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k}$.
Next,we calculate the cross product $\vec{AB} \times \vec{AC}$:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = \hat{i}(6-12) - \hat{j}(3-0) + \hat{k}(4-0) = -6\hat{i} - 3\hat{j} + 4\hat{k}$.
Now,find the magnitude of the cross product:
$|\vec{AB} \times \vec{AC}| = \sqrt{(-6)^2 + (-3)^2 + 4^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.
Finally,the area is $\frac{1}{2} \times \sqrt{61} = \frac{\sqrt{61}}{2}$ square units.

(D) We know that the cross products of unit vectors are $\hat{j} \times \hat{k} = \hat{i}$,$\hat{k} \times \hat{i} = \hat{j}$,and $\hat{i} \times \hat{j} = \hat{k}$.
Substituting these values into the expression,we get $\hat{i} \cdot \hat{i} + \hat{j} \cdot \hat{j} + \hat{k} \cdot \hat{k}$.
Since the dot product of a unit vector with itself is $1$ (i.e.,$\hat{i} \cdot \hat{i} = 1$,$\hat{j} \cdot \hat{j} = 1$,$\hat{k} \cdot \hat{k} = 1$),the expression becomes $1 + 1 + 1 = 3$.

(B) The direction vectors of the two lines are $\vec{b_1} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b_2} = 3\hat{i} + 2\hat{j} + 6\hat{k}$.
First,calculate the magnitudes of the direction vectors:
$|\vec{b_1}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$|\vec{b_2}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
Next,calculate the dot product of the direction vectors:
$\vec{b_1} \cdot \vec{b_2} = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$.
The angle $\theta$ between the lines is given by the formula $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
Substituting the values,we get $\cos \theta = \frac{19}{3 \cdot 7} = \frac{19}{21}$.
Therefore,$\theta = \cos^{-1}(\frac{19}{21})$.

(D) First,rewrite the lines in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
For the first line: $\frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{-2}$. The direction vector is $\vec{a} = (-3, \frac{2p}{7}, -2)$.
For the second line: $\frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$. The direction vector is $\vec{b} = (-\frac{3p}{7}, 1, -5)$.
Since the lines are perpendicular,their dot product must be zero: $\vec{a} \cdot \vec{b} = 0$.
$(-3)(-\frac{3p}{7}) + (\frac{2p}{7})(1) + (-2)(-5) = 0$.
$\frac{9p}{7} + \frac{2p}{7} + 10 = 0$.
$\frac{11p}{7} = -10$.
$p = -\frac{70}{11}$.

(C) The direction vectors of the two given lines are $\vec{v_1} = 3\hat{i} - 16\hat{j} + 7\hat{k}$ and $\vec{v_2} = 3\hat{i} + 8\hat{j} - 5\hat{k}$.
Since the required line is perpendicular to both lines,its direction vector $\vec{v}$ must be parallel to the cross product $\vec{v_1} \times \vec{v_2}$.
$\vec{v} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) = 24\hat{i} + 36\hat{j} + 72\hat{k}$.
Dividing by the common factor $12$,we get the direction vector as $2\hat{i} + 3\hat{j} + 6\hat{k}$.
The line passes through the point $(1, 2, -4)$,so its position vector is $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$.
The vector equation of the line is $\vec{r} = \vec{a} + \lambda\vec{v} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$.

(D) To find the minimum value of the objective function $z = 3x + 9y$,we evaluate $z$ at each corner point of the feasible region:
$1$. At $(0, 10)$: $z = 3(0) + 9(10) = 0 + 90 = 90$.
$2$. At $(5, 5)$: $z = 3(5) + 9(5) = 15 + 45 = 60$.
$3$. At $(15, 15)$: $z = 3(15) + 9(15) = 45 + 135 = 180$.
$4$. At $(0, 20)$: $z = 3(0) + 9(20) = 0 + 180 = 180$.
Comparing these values $(90, 60, 180, 180)$,the minimum value is $60$.

(A) Given the objective function $z = px + qy$.
At the corner point $(0, 10)$,$z = p(0) + q(10) = 90$.
This simplifies to $10q = 90$,which gives $q = 9$.
At the corner point $(5, 5)$,$z = p(5) + q(5) = 60$.
This simplifies to $5p + 5q = 60$,which reduces to $p + q = 12$.
Substituting the value $q = 9$ into the equation $p + q = 12$,we get $p + 9 = 12$,which implies $p = 3$.
Now,comparing $p = 3$ and $q = 9$,we observe that $9 = 3 \times 3$,which means $q = 3p$.

(B) We use the definition of conditional probability: $P(A'|B') = \frac{P(A' \cap B')}{P(B')}$.
By De Morgan's Law,$A' \cap B' = (A \cup B)'$.
Therefore,$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.
Given $P(A \cup B) = \frac{3}{11}$,we have $P(A' \cap B') = 1 - \frac{3}{11} = \frac{8}{11}$.
Also,$P(B') = 1 - P(B) = 1 - \frac{2}{11} = \frac{9}{11}$.
Substituting these values into the formula:
$P(A'|B') = \frac{8/11}{9/11} = \frac{8}{9}$.