MHT CET 2019 Mathematics Question Paper with Answer and Solution

(B) We have,$\frac{1-2(\cos 60^{\circ}-\cos 80^{\circ})}{2 \sin 10^{\circ}}$
$= \frac{1-2(\frac{1}{2}-\cos 80^{\circ})}{2 \sin 10^{\circ}}$
$= \frac{1-1+2 \cos 80^{\circ}}{2 \sin 10^{\circ}}$
$= \frac{2 \cos 80^{\circ}}{2 \sin 10^{\circ}}$
$= \frac{\cos(90^{\circ}-10^{\circ})}{\sin 10^{\circ}}$
$= \frac{\sin 10^{\circ}}{\sin 10^{\circ}} = 1$

(A) Let $I = \int \log x \cdot [\log (ex)]^{-2} dx$
Since $\log(ex) = \log e + \log x = 1 + \log x$,the integral becomes:
$I = \int \frac{\log x}{(1 + \log x)^{2}} dx$
Substitute $\log x = t$,which implies $x = e^{t}$ and $dx = e^{t} dt$.
$I = \int \frac{t}{(1 + t)^{2}} e^{t} dt$
We can rewrite the numerator as $(t + 1 - 1)$:
$I = \int e^{t} \left( \frac{t + 1 - 1}{(1 + t)^{2}} \right) dt$
$I = \int e^{t} \left( \frac{1}{1 + t} - \frac{1}{(1 + t)^{2}} \right) dt$
Using the standard integral formula $\int e^{x} [f(x) + f'(x)] dx = e^{x} f(x) + C$,where $f(t) = \frac{1}{1 + t}$ and $f'(t) = -\frac{1}{(1 + t)^{2}}$:
$I = e^{t} \cdot \frac{1}{1 + t} + C$
Substituting back $t = \log x$ and $e^{t} = x$:
$I = \frac{x}{1 + \log x} + C$

(D) Let $I = \int \frac{x^2+1}{x^4-x^2+1} \, dx$.
Divide the numerator and denominator by $x^2$:
$I = \int \frac{1 + \frac{1}{x^2}}{x^2 - 1 + \frac{1}{x^2}} \, dx$.
Rewrite the denominator as $(x - \frac{1}{x})^2 + 1$:
$I = \int \frac{1 + \frac{1}{x^2}}{(x - \frac{1}{x})^2 + 1} \, dx$.
Let $t = x - \frac{1}{x}$. Then $dt = (1 + \frac{1}{x^2}) \, dx$.
Substituting these into the integral:
$I = \int \frac{dt}{t^2 + 1} = \tan^{-1}(t) + c$.
Substituting $t = x - \frac{1}{x}$ back:
$I = \tan^{-1}\left(x - \frac{1}{x}\right) + c = \tan^{-1}\left(\frac{x^2-1}{x}\right) + c$.

(A) We have,$\tan 2 x = \tan ((x - \alpha) + (x + \alpha))$.
Using the identity $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get:
$\tan 2 x = \frac{\tan (x - \alpha) + \tan (x + \alpha)}{1 - \tan (x - \alpha) \tan (x + \alpha)}$.
Rearranging the terms:
$\tan 2 x (1 - \tan (x - \alpha) \tan (x + \alpha)) = \tan (x - \alpha) + \tan (x + \alpha)$.
$\tan 2 x - \tan (x - \alpha) \tan (x + \alpha) \tan 2 x = \tan (x - \alpha) + \tan (x + \alpha)$.
Thus,$\tan (x - \alpha) \tan (x + \alpha) \tan 2 x = \tan 2 x - \tan (x - \alpha) - \tan (x + \alpha)$.
Now,integrating both sides:
$\int \tan (x - \alpha) \tan (x + \alpha) \tan 2 x \ d x = \int (\tan 2 x - \tan (x - \alpha) - \tan (x + \alpha)) \ d x$.
$= \frac{1}{2} \log |\sec 2 x| - \log |\sec (x - \alpha)| - \log |\sec (x + \alpha)| + C$.
Comparing this with $p \log |\sec 2 x| + q \log |\sec (x + \alpha)| + r \log |\sec (x - \alpha)| + c$,we get:
$p = \frac{1}{2}$,$q = -1$,and $r = -1$.
Therefore,$p + q + r = \frac{1}{2} - 1 - 1 = \frac{1}{2} - 2 = \frac{-3}{2}$.

(A) Let $I = \int \frac{\cos x-\sin x}{8-\sin 2x} dx$.
We know that $\sin 2x = 2 \sin x \cos x$ and $1 = \sin^2 x + \cos^2 x$.
So,$8 - \sin 2x = 9 - (1 + 2 \sin x \cos x) = 9 - (\sin x + \cos x)^2$.
Thus,$I = \int \frac{\cos x - \sin x}{3^2 - (\sin x + \cos x)^2} dx$.
Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$.
Substituting these into the integral,we get $I = \int \frac{dt}{3^2 - t^2}$.
Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C$,we have:
$I = \frac{1}{2(3)} \log \left| \frac{3+t}{3-t} \right| + C = \frac{1}{6} \log \left| \frac{3+\sin x + \cos x}{3 - (\sin x + \cos x)} \right| + C$.
Comparing this with the given expression $\frac{1}{p} \log \left[\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right]+c$,we find $p = 6$.

(B) Let $I = \int \frac{dx}{(x^2+1)^2}$.
Substitute $x = \tan \theta$,then $dx = \sec^2 \theta d \theta$.
$I = \int \frac{\sec^2 \theta d \theta}{(\tan^2 \theta + 1)^2} = \int \frac{\sec^2 \theta d \theta}{\sec^4 \theta} = \int \cos^2 \theta d \theta$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2 \theta}{2}$,we get:
$I = \frac{1}{2} \int (1 + \cos 2 \theta) d \theta = \frac{1}{2} (\theta + \frac{\sin 2 \theta}{2}) + c$.
Since $\sin 2 \theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}$ and $\theta = \tan^{-1} x$,we have:
$I = \frac{1}{2} \tan^{-1} x + \frac{1}{4} \cdot \frac{2x}{1 + x^2} + c = \frac{1}{2} \tan^{-1} x + \frac{x}{2(1 + x^2)} + c$.

(B) Let $I = \int \frac{\cos x + x \sin x}{x(x + \cos x)} dx$.
Observe the denominator $f(x) = x^2 + x \cos x$.
The derivative of the denominator is $f'(x) = \frac{d}{dx}(x^2 + x \cos x) = 2x + \cos x - x \sin x$. This does not match the numerator directly.
Let us rewrite the integral as:
$I = \int \frac{x + \cos x + x \sin x - x}{x(x + \cos x)} dx$
$I = \int \left( \frac{x + \cos x}{x(x + \cos x)} + \frac{x \sin x - x}{x(x + \cos x)} \right) dx$
$I = \int \left( \frac{1}{x} + \frac{x(\sin x - 1)}{x(x + \cos x)} \right) dx$
$I = \int \frac{1}{x} dx + \int \frac{\sin x - 1}{x + \cos x} dx$
Let $u = x + \cos x$,then $du = (1 - \sin x) dx$,which means $-(1 - \sin x) dx = du$,or $(\sin x - 1) dx = -du$.
$I = \ln|x| - \int \frac{1}{u} du$
$I = \ln|x| - \ln|u| + c$
$I = \ln|x| - \ln|x + \cos x| + c$
$I = \ln \left| \frac{x}{x + \cos x} \right| + c$.

(A) Let $I = \int \frac{\sqrt{x^2-a^2}}{x} d x$.
Substitute $x = a \sec \theta$,then $d x = a \sec \theta \tan \theta d \theta$.
Substituting these into the integral:
$I = \int \frac{\sqrt{a^2 \sec^2 \theta - a^2}}{a \sec \theta} (a \sec \theta \tan \theta) d \theta$
$I = \int \frac{a \tan \theta}{a \sec \theta} (a \sec \theta \tan \theta) d \theta$
$I = a \int \tan^2 \theta d \theta$
$I = a \int (\sec^2 \theta - 1) d \theta$
$I = a (\tan \theta - \theta) + c$
Since $x = a \sec \theta$,we have $\sec \theta = \frac{x}{a}$,so $\theta = \sec^{-1}(\frac{x}{a})$ and $\tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{(\frac{x}{a})^2 - 1} = \frac{\sqrt{x^2-a^2}}{a}$.
Substituting back:
$I = a (\frac{\sqrt{x^2-a^2}}{a} - \sec^{-1}(\frac{x}{a})) + c$
$I = \sqrt{x^2-a^2} - a \sec^{-1}(\frac{x}{a}) + c$.

(C) Let $I = \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{2\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ . . . $(i)$
Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a = \frac{\pi}{18}$ and $b = \frac{4\pi}{9}$,we have $a+b = \frac{\pi}{18} + \frac{8\pi}{18} = \frac{9\pi}{18} = \frac{\pi}{2}$.
So,$I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} dx$
$I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ . . . $(ii)$
Adding $(i)$ and $(ii)$:
$2I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) dx$
$2I = 2 \int_{\frac{\pi}{18}}^{\frac{4\pi}{9}} 1 dx = 2 [x]_{\frac{\pi}{18}}^{\frac{4\pi}{9}}$
$2I = 2 \left( \frac{4\pi}{9} - \frac{\pi}{18} \right) = 2 \left( \frac{8\pi - \pi}{18} \right) = 2 \left( \frac{7\pi}{18} \right) = \frac{7\pi}{9}$
$I = \frac{7\pi}{18}$

(C) Given equation: $4 \sin ^{-1} x + 6 \cos ^{-1} x = 3 \pi$.
We know the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,which implies $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$.
Substituting this into the equation:
$4 \sin ^{-1} x + 6(\frac{\pi}{2} - \sin ^{-1} x) = 3 \pi$
$4 \sin ^{-1} x + 3 \pi - 6 \sin ^{-1} x = 3 \pi$
$-2 \sin ^{-1} x + 3 \pi = 3 \pi$
$-2 \sin ^{-1} x = 0$
$\sin ^{-1} x = 0$
$x = \sin(0) = 0$.

(B) Let $L = \tan ^{-1} \frac{1}{3} + \tan ^{-1} \frac{1}{5} + \tan ^{-1} \frac{1}{7} + \tan ^{-1} \frac{1}{8}$.
Using the formula $\tan ^{-1} x + \tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$:
First,group the terms: $L = \left( \tan ^{-1} \frac{1}{3} + \tan ^{-1} \frac{1}{5} \right) + \left( \tan ^{-1} \frac{1}{7} + \tan ^{-1} \frac{1}{8} \right)$.
Calculate the first part: $\tan ^{-1} \left( \frac{1/3 + 1/5}{1 - (1/3)(1/5)} \right) = \tan ^{-1} \left( \frac{8/15}{14/15} \right) = \tan ^{-1} \left( \frac{8}{14} \right) = \tan ^{-1} \left( \frac{4}{7} \right)$.
Calculate the second part: $\tan ^{-1} \left( \frac{1/7 + 1/8}{1 - (1/7)(1/8)} \right) = \tan ^{-1} \left( \frac{15/56}{55/56} \right) = \tan ^{-1} \left( \frac{15}{55} \right) = \tan ^{-1} \left( \frac{3}{11} \right)$.
Now combine them: $L = \tan ^{-1} \left( \frac{4}{7} \right) + \tan ^{-1} \left( \frac{3}{11} \right) = \tan ^{-1} \left( \frac{4/7 + 3/11}{1 - (4/7)(3/11)} \right)$.
$L = \tan ^{-1} \left( \frac{44/77 + 21/77}{1 - 12/77} \right) = \tan ^{-1} \left( \frac{65/77}{65/77} \right) = \tan ^{-1} (1) = \frac{\pi}{4}$.

(C) Given,$y = \tan^{-1} \left( \frac{1 - \cos 3x}{\sin 3x} \right)$.
Using trigonometric identities $1 - \cos \theta = 2 \sin^2 \left( \frac{\theta}{2} \right)$ and $\sin \theta = 2 \sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right)$,we have:
$y = \tan^{-1} \left( \frac{2 \sin^2 \left( \frac{3x}{2} \right)}{2 \sin \left( \frac{3x}{2} \right) \cos \left( \frac{3x}{2} \right)} \right)$
$y = \tan^{-1} \left( \frac{\sin \left( \frac{3x}{2} \right)}{\cos \left( \frac{3x}{2} \right)} \right)$
$y = \tan^{-1} \left( \tan \left( \frac{3x}{2} \right) \right)$
$y = \frac{3x}{2}$
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{3x}{2} \right) = \frac{3}{2}$.

(A) Given that $f(x)$ is continuous at $x = 0$,we have $\lim_{x \to 0} f(x) = f(0) = k$.
Since the limit is of the form $1^\infty$,we use the formula $\lim_{x \to a} [g(x)]^{h(x)} = e^{\lim_{x \to a} [g(x) - 1]h(x)}$.
Here,$g(x) = \tan(\frac{\pi}{4} + x)$ and $h(x) = \frac{1}{x}$.
$k = e^{\lim_{x \to 0} [\tan(\frac{\pi}{4} + x) - 1] \cdot \frac{1}{x}}$.
Using $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we get $\tan(\frac{\pi}{4} + x) = \frac{1 + \tan x}{1 - \tan x}$.
$k = e^{\lim_{x \to 0} [\frac{1 + \tan x}{1 - \tan x} - 1] \cdot \frac{1}{x}} = e^{\lim_{x \to 0} [\frac{1 + \tan x - 1 + \tan x}{1 - \tan x}] \cdot \frac{1}{x}}$.
$k = e^{\lim_{x \to 0} \frac{2 \tan x}{x(1 - \tan x)}} = e^{2 \cdot \lim_{x \to 0} \frac{\tan x}{x} \cdot \lim_{x \to 0} \frac{1}{1 - \tan x}}$.
Since $\lim_{x \to 0} \frac{\tan x}{x} = 1$ and $\lim_{x \to 0} \frac{1}{1 - \tan x} = 1$,we have $k = e^{2 \cdot 1 \cdot 1} = e^2$.

(B) Given $A = \left[ \begin{array}{cc} 1+2i & i \\ -i & 1-2i \end{array} \right]$.
We know that for any square matrix $A$,$A(\text{adj } A) = |A|I$,where $I$ is the identity matrix.
First,calculate the determinant $|A|$:
$|A| = (1+2i)(1-2i) - (i)(-i)$
$|A| = (1^2 - (2i)^2) - (-i^2)$
Since $i^2 = -1$,we have:
$|A| = (1 - 4(-1)) - (-(-1))$
$|A| = (1 + 4) - 1$
$|A| = 5 - 1 = 4$.
Therefore,$A(\text{adj } A) = |A|I = 4I$.

(A) Given that $\omega$ is a complex cube root of unity,we have $\omega^3 = 1$.
The inverse of a diagonal matrix $A = \text{diag}(a, b, c)$ is given by $A^{-1} = \text{diag}(a^{-1}, b^{-1}, c^{-1})$.
Here,$A = \begin{bmatrix} \omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Therefore,$A^{-1} = \begin{bmatrix} \omega^{-1} & 0 & 0 \\ 0 & (\omega^2)^{-1} & 0 \\ 0 & 0 & 1^{-1} \end{bmatrix} = \begin{bmatrix} \frac{1}{\omega} & 0 & 0 \\ 0 & \frac{1}{\omega^2} & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Since $\omega^3 = 1$,we have $\frac{1}{\omega} = \omega^2$ and $\frac{1}{\omega^2} = \omega$.
Thus,$A^{-1} = \begin{bmatrix} \omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

(A) Given that $A$ is a non-singular matrix,so $|A| \neq 0$.
Given equation: $(A+I)(A-I) = 0$.
Expanding the product: $A^2 - AI + IA - I^2 = 0$.
Since $AI = IA = A$ and $I^2 = I$,we get $A^2 - I = 0$.
Therefore,$A^2 = I$.
Multiplying both sides by $A^{-1}$,we get $A^{-1}A^2 = A^{-1}I$.
This simplifies to $A = A^{-1}$.
Now,substituting $A^{-1} = A$ into the expression $A + A^{-1}$,we get $A + A = 2A$.

(A) Given $A = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}$.
We know that $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.
First,calculate the determinant $|A| = (x \times 0) - (1 \times 1) = -1$.
Next,find the adjoint of $A$ by swapping the diagonal elements and changing the signs of the off-diagonal elements:
$\text{adj}(A) = \begin{bmatrix} 0 & -1 \\ -1 & x \end{bmatrix}$.
Therefore,$A^{-1} = \frac{1}{-1} \begin{bmatrix} 0 & -1 \\ -1 & x \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -x \end{bmatrix}$.
Given $A = A^{-1}$,we equate the matrices:
$\begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -x \end{bmatrix}$.
Comparing the corresponding elements,we get $x = 0$ and $0 = -x$,which implies $x = 0$.

(D) Given that $A$ is a non-singular matrix,so $|A| \neq 0$.
Given the equation $(A-2I)(A-4I) = 0$.
Expanding the product,we get $A^2 - 4A - 2A + 8I = 0$,which simplifies to $A^2 - 6A + 8I = 0$.
Since $A$ is non-singular,$A^{-1}$ exists. Multiplying the entire equation by $A^{-1}$ on both sides:
$A^{-1}(A^2 - 6A + 8I) = A^{-1}(0)$
$A - 6I + 8A^{-1} = 0$
Rearranging the terms to solve for $A + 8A^{-1}$,we get:
$A + 8A^{-1} = 6I$.

(D) Let $p$ be the probability of a case being solved,so $p = 25\% = \frac{1}{4}$.
Let $q$ be the probability of a case not being solved,so $q = 1 - p = \frac{3}{4}$.
Given $n = 6$ trials,we use the binomial distribution formula $P(X=r) = {^nC_r} p^r q^{n-r}$.
We need the probability that at least $5$ cases are solved,which is $P(X \ge 5) = P(X=5) + P(X=6)$.
$P(X=5) = {^6C_5} \left(\frac{1}{4}\right)^5 \left(\frac{3}{4}\right)^1 = 6 \times \frac{1}{1024} \times \frac{3}{4} = \frac{18}{4096}$.
$P(X=6) = {^6C_6} \left(\frac{1}{4}\right)^6 \left(\frac{3}{4}\right)^0 = 1 \times \frac{1}{4096} \times 1 = \frac{1}{4096}$.
Therefore,$P(X \ge 5) = \frac{18}{4096} + \frac{1}{4096} = \frac{19}{4096}$.

(A) The sample space for tossing $2$ coins is $\{HH, HT, TH, TT\}$.
The probabilities are:
$P(X=5) = P(HH) = \frac{1}{4}$
$P(X=2) = P(HT, TH) = \frac{2}{4} = \frac{1}{2}$
$P(X=1) = P(TT) = \frac{1}{4}$
We calculate the mean $E(X) = \sum p_i x_i = (5 \times \frac{1}{4}) + (2 \times \frac{1}{2}) + (1 \times \frac{1}{4}) = \frac{5}{4} + 1 + \frac{1}{4} = \frac{6}{4} + 1 = \frac{3}{2} + 1 = \frac{5}{2}$.
Next,we calculate $E(X^2) = \sum p_i x_i^2 = (5^2 \times \frac{1}{4}) + (2^2 \times \frac{1}{2}) + (1^2 \times \frac{1}{4}) = \frac{25}{4} + 2 + \frac{1}{4} = \frac{26}{4} + 2 = \frac{13}{2} + 2 = \frac{17}{2}$.
Variance $Var(X) = E(X^2) - [E(X)]^2 = \frac{17}{2} - (\frac{5}{2})^2 = \frac{17}{2} - \frac{25}{4} = \frac{34-25}{4} = \frac{9}{4}$.

(A) Given the p.d.f. of a random variable $X$: $f(x) = \begin{cases} 3(1 - 2x^2), & 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$
The probability is calculated as: $P\left(\frac{1}{4} < X < \frac{1}{3}\right) = \int_{1/4}^{1/3} 3(1 - 2x^2) \, dx$
$= 3 \left[ x - \frac{2}{3}x^3 \right]_{1/4}^{1/3}$
$= 3 \left[ \left( \frac{1}{3} - \frac{2}{3} \cdot \frac{1}{27} \right) - \left( \frac{1}{4} - \frac{2}{3} \cdot \frac{1}{64} \right) \right]$
$= 3 \left[ \left( \frac{1}{3} - \frac{2}{81} \right) - \left( \frac{1}{4} - \frac{1}{96} \right) \right]$
$= 3 \left[ \frac{27-2}{81} - \frac{24-1}{96} \right] = 3 \left[ \frac{25}{81} - \frac{23}{96} \right]$
$= 3 \left[ \frac{25 \times 32 - 23 \times 27}{2592} \right] = 3 \left[ \frac{800 - 621}{2592} \right]$
$= 3 \times \frac{179}{2592} = \frac{179}{864}$

(A) We have $p = \frac{3}{4}$,so $q = 1 - p = \frac{1}{4}$.
It is given that $P(X \ge 1) \ge 0.9375$.
Using the complement rule,$P(X \ge 1) = 1 - P(X = 0)$.
Thus,$1 - P(X = 0) \ge 0.9375$.
$1 - ^nC_0 (p^0) (q)^n \ge 0.9375$.
$1 - (\frac{1}{4})^n \ge 0.9375$.
$1 - 0.9375 \ge (\frac{1}{4})^n$.
$0.0625 \ge (\frac{1}{4})^n$.
Since $0.0625 = \frac{625}{10000} = \frac{1}{16}$,we have $\frac{1}{16} \ge (\frac{1}{4})^n$.
$(\frac{1}{4})^2 \ge (\frac{1}{4})^n$.
Since the base is less than $1$,the inequality reverses for the exponents: $n \ge 2$.
The least value of $n$ is $2$.

(B) Key Idea: Use $p + q = 1$ and $Var(X) = E(X^2) - (E(X))^2$.
Given that the standard deviation of $X$ is $\sqrt{3pq}$,the variance is $Var(X) = (\sqrt{3pq})^2 = 3pq$.
Given that the mean $E(X) = 3p$.
Using the formula $Var(X) = E(X^2) - (E(X))^2$,we have:
$3pq = E(X^2) - (3p)^2$
$E(X^2) = 3pq + 9p^2$.
Since $p + q = 1$,we can substitute $q = 1 - p$:
$E(X^2) = 3p(1 - p) + 9p^2$
$E(X^2) = 3p - 3p^2 + 9p^2$
$E(X^2) = 3p + 6p^2$
$E(X^2) = 3p(1 + 2p)$.

(A) Given the p.d.f. of a random variable $x$ is $f(x) = \frac{1}{4a}$ for $0 < x < 4a$.
We are given the equation $P(x < \frac{3a}{2}) = k P(x > \frac{5a}{2})$.
Calculating the left side: $P(x < \frac{3a}{2}) = \int_{0}^{\frac{3a}{2}} \frac{1}{4a} dx = \frac{1}{4a} [x]_{0}^{\frac{3a}{2}} = \frac{1}{4a} \times \frac{3a}{2} = \frac{3}{8}$.
Calculating the right side: $P(x > \frac{5a}{2}) = \int_{\frac{5a}{2}}^{4a} \frac{1}{4a} dx = \frac{1}{4a} [x]_{\frac{5a}{2}}^{4a} = \frac{1}{4a} (4a - \frac{5a}{2}) = \frac{1}{4a} \times \frac{3a}{2} = \frac{3}{8}$.
Substituting these values into the given equation: $\frac{3}{8} = k \times \frac{3}{8}$.
Therefore,$k = 1$.

(B) We know that for a binomial distribution,the mean is given by $\mu = np = 18$ and the variance is given by $\sigma^2 = npq = 12$.
Dividing the variance by the mean,we get: $\frac{npq}{np} = \frac{12}{18}$.
This simplifies to $q = \frac{2}{3}$.
Since $p + q = 1$,we have $p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$.

(A) The probability $P(a \leq x \leq b)$ is given by $F(b) - F(a)$.
Given $F(x) = \frac{x-25}{10}$.
We need to find $P(27 \leq x \leq 33) = F(33) - F(27)$.
$F(33) = \frac{33-25}{10} = \frac{8}{10} = 0.8$.
$F(27) = \frac{27-25}{10} = \frac{2}{10} = 0.2$.
Therefore,$P(27 \leq x \leq 33) = 0.8 - 0.2 = 0.6 = \frac{3}{5}$.

(A) For any probability distribution,the sum of all probabilities must be equal to $1$.
$\sum P(X=x) = K + 3K + 5K + 7K + 8K + K = 1$
$25K = 1 \implies K = \frac{1}{25}$
We need to find $P(2 \leq X < 5)$,which includes the values $X = 2, 3, 4$.
$P(2 \leq X < 5) = P(X=2) + P(X=3) + P(X=4)$
$= 3K + 5K + 7K = 15K$
Substituting $K = \frac{1}{25}$:
$= 15 \times \frac{1}{25} = \frac{15}{25} = \frac{3}{5}$

(B) The objective function is $z=6x+8y$. The constraints are $x-y \geq 0$,$x+3y \leq 12$,$x \geq 0$,and $y \geq 0$.
To find the feasible region,we plot the lines $x-y=0$ and $x+3y=12$.
The intersection point of $x-y=0$ and $x+3y=12$ is found by substituting $x=y$ into $x+3y=12$,which gives $4y=12$,so $y=3$ and $x=3$. Thus,the intersection point is $B(3, 3)$.
The feasible region is a triangle with vertices $O(0, 0)$,$A(0, 4)$ (from $x+3y=12$ when $x=0$),and $B(3, 3)$.
Evaluating $z=6x+8y$ at these corner points:
At $O(0, 0)$: $z = 6(0) + 8(0) = 0$.
At $A(0, 4)$: $z = 6(0) + 8(4) = 32$.
At $B(3, 3)$: $z = 6(3) + 8(3) = 18 + 24 = 42$.
The maximum value is $42$.

(D) Given that $c = (x-2)a + b$ and $d = (2x+1)a - b$ are collinear vectors.
Since $a$ and $b$ are non-collinear,we can write $c = \lambda d$ for some scalar $\lambda$.
$(x-2)a + b = \lambda((2x+1)a - b)$
$(x-2)a + b = \lambda(2x+1)a - \lambda b$
Comparing the coefficients of $a$ and $b$,we get:
$x-2 = \lambda(2x+1)$ and $1 = -\lambda$.
From the second equation,$\lambda = -1$.
Substituting $\lambda = -1$ into the first equation:
$x-2 = -1(2x+1)$
$x-2 = -2x-1$
$3x = 1$
$x = \frac{1}{3}$

(C) The direction cosines $l, m, n$ of a line must satisfy the condition $l^2 + m^2 + n^2 = 1$.
For option $A$: $(\sqrt{\frac{1}{5}})^2 + (-\sqrt{\frac{1}{2}})^2 + (\sqrt{\frac{3}{10}})^2 = \frac{1}{5} + \frac{1}{2} + \frac{3}{10} = \frac{2+5+3}{10} = \frac{10}{10} = 1$.
For option $B$: $(\frac{1}{\sqrt{3}})^2 + (\frac{-1}{\sqrt{3}})^2 + (\frac{1}{\sqrt{3}})^2 = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$.
For option $C$: $(\frac{1}{\sqrt{2}})^2 + (\frac{-1}{\sqrt{2}})^2 + (\frac{-1}{\sqrt{2}})^2 = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \neq 1$.
For option $D$: $(\frac{1}{\sqrt{2}})^2 + (\frac{-1}{\sqrt{2}})^2 + 0^2 = \frac{1}{2} + \frac{1}{2} + 0 = 1$.
Since the sum of squares for option $C$ is not $1$,it cannot be the direction cosines of a line.

(D) The equation of the family of planes passing through the line of intersection of the planes $x+2y+3z-4=0$ and $4x+3y+2z-1=0$ is given by $(x+2y+3z-4) + \lambda(4x+3y+2z-1) = 0$.
Since the plane passes through the origin $(0, 0, 0)$,we substitute $x=0, y=0, z=0$ into the equation:
$(0+0+0-4) + \lambda(0+0+0-1) = 0
\Rightarrow -4 - \lambda = 0
\Rightarrow \lambda = -4$.
Substituting $\lambda = -4$ back into the family equation:
$(x+2y+3z-4) - 4(4x+3y+2z-1) = 0
\Rightarrow x+2y+3z-4 - 16x - 12y - 8z + 4 = 0
\Rightarrow -15x - 10y - 5z = 0
\Rightarrow 3x + 2y + z = 0$.
The direction ratios of the normal to the plane $ax+by+cz+d=0$ are $(a, b, c)$.
Thus,the direction ratios of the normal to the plane $3x+2y+z=0$ are $3, 2, 1$.

(D) Key Idea: If lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ are perpendicular,then $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Given lines are:
$\frac{2x-4}{\lambda} = \frac{y-1}{2} = \frac{z-3}{1} \Rightarrow \frac{x-2}{\lambda/2} = \frac{y-1}{2} = \frac{z-3}{1}$ $(i)$
And $\frac{x-1}{1} = \frac{3y-1}{\lambda} = \frac{z-2}{1} \Rightarrow \frac{x-1}{1} = \frac{y-1/3}{\lambda/3} = \frac{z-2}{1}$ (ii)
Since lines $(i)$ and (ii) are perpendicular,the dot product of their direction ratios must be zero:
$(\frac{\lambda}{2})(1) + (2)(\frac{\lambda}{3}) + (1)(1) = 0$
$\frac{\lambda}{2} + \frac{2\lambda}{3} + 1 = 0$
Multiplying by $6$ to clear the denominators:
$3\lambda + 4\lambda + 6 = 0$
$7\lambda = -6$
$\lambda = -\frac{6}{7}$

(A) Given that $\triangle PQR$ is right-angled at $Q$.
Therefore,the vectors $\vec{QP}$ and $\vec{QR}$ are perpendicular to each other,which means their dot product is zero: $\vec{QP} \cdot \vec{QR} = 0$.
First,find the vectors $\vec{QP}$ and $\vec{QR}$:
$\vec{QP} = (6-1, 10-0, 10-(-5)) = (5, 10, 15)$
$\vec{QR} = (6-1, -10-0, \lambda-(-5)) = (5, -10, \lambda+5)$
Now,calculate the dot product:
$\vec{QP} \cdot \vec{QR} = (5)(5) + (10)(-10) + (15)(\lambda+5) = 0$
$25 - 100 + 15\lambda + 75 = 0$
$-75 + 15\lambda + 75 = 0$
$15\lambda = 0$
$\lambda = 0$

(C) Let the given lines be:
$L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = k_1$
$L_2: \frac{x-3}{1}=\frac{y-\lambda}{2}=\frac{z}{1} = k_2$
Any point on $L_1$ is $(2k_1+1, 3k_1-1, 4k_1+1)$ and any point on $L_2$ is $(k_2+3, 2k_2+\lambda, k_2)$.
If the lines intersect,there exist $k_1, k_2$ such that:
$2k_1+1 = k_2+3 \Rightarrow 2k_1 - k_2 = 2$ $(i)$
$4k_1+1 = k_2 \Rightarrow 4k_1 - k_2 = -1$ $(ii)$
Subtracting $(i)$ from $(ii)$:
$(4k_1 - k_2) - (2k_1 - k_2) = -1 - 2$
$2k_1 = -3 \Rightarrow k_1 = -\frac{3}{2}$
Substituting $k_1$ in $(ii)$:
$4(-\frac{3}{2}) - k_2 = -1 \Rightarrow -6 - k_2 = -1 \Rightarrow k_2 = -5$
Now,equate the $y$-coordinates:
$3k_1 - 1 = 2k_2 + \lambda$
$3(-\frac{3}{2}) - 1 = 2(-5) + \lambda$
$-\frac{9}{2} - 1 = -10 + \lambda$
$-\frac{11}{2} = -10 + \lambda$
$\lambda = 10 - \frac{11}{2} = \frac{20-11}{2} = \frac{9}{2}$

(C) Since the points $P(6, -1, 2)$,$Q(8, -7, 2\lambda)$,and $R(5, 2, 4)$ are collinear,the direction ratios of the line segments $PQ$ and $PR$ must be proportional.
The direction ratios of $PQ$ are $(8-6, -7-(-1), 2\lambda-2) = (2, -6, 2\lambda-2)$.
The direction ratios of $PR$ are $(5-6, 2-(-1), 4-2) = (-1, 3, 2)$.
Since the points are collinear,the ratio of the direction ratios must be equal:
$\frac{2}{-1} = \frac{-6}{3} = \frac{2\lambda-2}{2}$
$-2 = -2 = \lambda-1$
Equating the last part: $-2 = \lambda-1$,which gives $\lambda = -1$.

(D) The direction ratios of the first line are $\vec{a_1} = (2, -2, 1)$.
The direction ratios of the second line are $\vec{a_2} = (1, 2, 2)$.
The angle $\theta$ between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values:
$\cos \theta = \frac{|(2)(1) + (-2)(2) + (1)(2)|}{\sqrt{2^2 + (-2)^2 + 1^2} \sqrt{1^2 + 2^2 + 2^2}}$
$\cos \theta = \frac{|2 - 4 + 2|}{\sqrt{4 + 4 + 1} \sqrt{1 + 4 + 4}}$
$\cos \theta = \frac{0}{\sqrt{9} \sqrt{9}} = 0$
Since $\cos \theta = 0$,we have $\theta = 90^{\circ}$.

(B) The foot of the perpendicular drawn from the origin $O(0, 0, 0)$ to the plane is given as $P(4, -2, -5)$.
Since the line segment $OP$ is perpendicular to the plane,the vector $\vec{OP}$ acts as the normal vector $\vec{n}$ to the plane.
Thus,the direction ratios of the normal are $(4 - 0, -2 - 0, -5 - 0) = (4, -2, -5)$.
The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $(a, b, c)$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.
Substituting the values,we get $4(x - 4) - 2(y + 2) - 5(z + 5) = 0$.
Expanding this,we have $4x - 16 - 2y - 4 - 5z - 25 = 0$.
Simplifying,we get $4x - 2y - 5z - 45 = 0$,or $4x - 2y - 5z = 45$.

(D) The equation of a plane parallel to $x-2y+2z+4=0$ is of the form $x-2y+2z+k=0$.
The distance $d$ from a point $(x_1, y_1, z_1)$ to the plane $Ax+By+Cz+D=0$ is given by $d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$.
Given the point $(1, 2, 3)$ and distance $d=1$,we have:
$1 = \frac{|1(1)-2(2)+2(3)+k|}{\sqrt{1^2+(-2)^2+2^2}}$
$1 = \frac{|1-4+6+k|}{\sqrt{1+4+4}}$
$1 = \frac{|3+k|}{\sqrt{9}}$
$|3+k| = 3$
This implies $3+k = 3$ or $3+k = -3$.
Case $1$: $3+k = 3 \Rightarrow k = 0$. The equation is $x-2y+2z=0$.
Case $2$: $3+k = -3 \Rightarrow k = -6$. The equation is $x-2y+2z-6=0$.
Thus,the required equations are $x-2y+2z=0$ and $x-2y+2z-6=0$.

(A) The direction ratios of the line joining the points $(-3, 1, 2)$ and $(2, 3, 4)$ are $(2 - (-3)), (3 - 1), (4 - 2)$,which are $5, 2, 2$.
Since the plane is perpendicular to this line,the normal vector to the plane is $\vec{n} = 5\hat{i} + 2\hat{j} + 2\hat{k}$.
The equation of a plane passing through a point $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$ with normal $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.
Substituting the values,we get $(\vec{r} - (-\hat{i} + 2\hat{j} + \hat{k})) \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = 0$.
This simplifies to $\vec{r} \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = (-\hat{i} + 2\hat{j} + \hat{k}) \cdot (5\hat{i} + 2\hat{j} + 2\hat{k})$.
Calculating the dot product on the right side: $(-1)(5) + (2)(2) + (1)(2) = -5 + 4 + 2 = 1$.
Thus,the equation of the plane is $\vec{r} \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = 1$.

(D) The coordinates of the foot of the perpendicular $(x, y, z)$ from a point $(x_1, y_1, z_1)$ to the plane $ax + by + cz + d = 0$ are given by the formula: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = -\frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$.
Given the plane $2x - y + 5z - 3 = 0$ and the origin $(0, 0, 0)$,we have $x_1 = 0, y_1 = 0, z_1 = 0$.
Substituting these values into the formula:
$\frac{x - 0}{2} = \frac{y - 0}{-1} = \frac{z - 0}{5} = -\frac{2(0) - 1(0) + 5(0) - 3}{2^2 + (-1)^2 + 5^2}$.
$\frac{x}{2} = \frac{y}{-1} = \frac{z}{5} = -\frac{-3}{4 + 1 + 25} = \frac{3}{30} = \frac{1}{10}$.
Equating each part to $\frac{1}{10}$:
$x = 2 \times \frac{1}{10} = \frac{1}{5}$.
$y = -1 \times \frac{1}{10} = -\frac{1}{10}$.
$z = 5 \times \frac{1}{10} = \frac{1}{2}$.
Thus,the coordinates are $\left(\frac{1}{5}, -\frac{1}{10}, \frac{1}{2}\right)$.

(C) Let $E = \sin \left[3 \sin ^{-1}(0.4)\right]$.
Put $\sin ^{-1}(0.4) = \theta$,which implies $\sin \theta = 0.4$.
Using the identity $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$:
$E = 3(0.4) - 4(0.4)^3$
$E = 1.2 - 4(0.064)$
$E = 1.2 - 0.256$
$E = 0.944$

(B) Two vectors $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$ are collinear if their components are proportional,i.e.,$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = k$.
Given vectors are $x \hat{i}-3 \hat{j}+7 \hat{k}$ and $\hat{i}+y \hat{j}-z \hat{k}$.
Therefore,$\frac{x}{1} = \frac{-3}{y} = \frac{7}{-z} = k$.
From this,we get $x = k$,$y = -\frac{3}{k}$,and $z = -\frac{7}{k}$.
Now,substitute these values into the expression $\frac{x y^2}{z}$:
$\frac{x y^2}{z} = \frac{k \cdot (-\frac{3}{k})^2}{-\frac{7}{k}} = \frac{k \cdot \frac{9}{k^2}}{-\frac{7}{k}} = \frac{\frac{9}{k}}{-\frac{7}{k}} = -\frac{9}{7}$.

(A) We have the expression $a \cdot[(b+c) \times(a+b+c)]$.
Using the distributive property of the cross product,we expand the term inside the bracket:
$(b+c) \times(a+b+c) = (b \times a) + (b \times b) + (b \times c) + (c \times a) + (c \times b) + (c \times c)$.
Since the cross product of any vector with itself is zero ($b \times b = 0$ and $c \times c = 0$) and $c \times b = -(b \times c)$,the expression simplifies to:
$(b \times a) + (b \times c) + (c \times a) - (b \times c) = (b \times a) + (c \times a)$.
Now,taking the dot product with $a$:
$a \cdot [(b \times a) + (c \times a)] = a \cdot (b \times a) + a \cdot (c \times a)$.
By the definition of the scalar triple product,$a \cdot (b \times a) = [a \ b \ a]$ and $a \cdot (c \times a) = [a \ c \ a]$.
Since any scalar triple product with two identical vectors is zero,we have $[a \ b \ a] = 0$ and $[a \ c \ a] = 0$.
Therefore,the final result is $0 + 0 = 0$.

(B) The scalar triple product of vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the determinant of the matrix formed by their components: $[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$.
Given the vectors $-3 \hat{i}+7 \hat{j}-3 \hat{k}$,$3 \hat{i}-7 \hat{j}+\lambda \hat{k}$,and $7 \hat{i}-5 \hat{j}-3 \hat{k}$,their scalar triple product is $272$.
$\begin{vmatrix} -3 & 7 & -3 \\ 3 & -7 & \lambda \\ 7 & -5 & -3 \end{vmatrix} = 272$
Expanding the determinant along the first row:
$-3((-7)(-3) - (-5)(\lambda)) - 7((3)(-3) - (7)(\lambda)) - 3((3)(-5) - (7)(-7)) = 272$
$-3(21 + 5\lambda) - 7(-9 - 7\lambda) - 3(-15 + 49) = 272$
$-63 - 15\lambda + 63 + 49\lambda - 3(34) = 272$
$34\lambda - 102 = 272$
$34\lambda = 374$
$\lambda = \frac{374}{34} = 11$.

(A) The given equation of the plane is $r = (2 \hat{i} + \hat{k}) + \lambda(\hat{i}) + \mu(\hat{i} + 2 \hat{j} - 3 \hat{k})$.
This plane passes through the point $a = 2 \hat{i} + \hat{k}$ and is parallel to the vectors $b = \hat{i}$ and $c = \hat{i} + 2 \hat{j} - 3 \hat{k}$.
The scalar product form of the plane is $r \cdot (b \times c) = a \cdot (b \times c)$.
First,calculate the normal vector $n = b \times c$:
$n = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 1 & 2 & -3 \end{vmatrix} = \hat{i}(0 - 0) - \hat{j}(-3 - 0) + \hat{k}(2 - 0) = 3 \hat{j} + 2 \hat{k}$.
Wait,re-calculating the cross product: $\hat{i}(0) - \hat{j}(-3) + \hat{k}(2) = 3 \hat{j} + 2 \hat{k}$.
Actually,the question specifies $r \cdot (3 \hat{i} + 2 \hat{k}) = \alpha$. Let us re-check the cross product: $b = \hat{i} = (1, 0, 0)$,$c = (1, 2, -3)$.
$b \times c = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 1 & 2 & -3 \end{vmatrix} = \hat{i}(0) - \hat{j}(-3) + \hat{k}(2) = 3 \hat{j} + 2 \hat{k}$.
There seems to be a typo in the provided question's normal vector. Assuming the normal is $3 \hat{j} + 2 \hat{k}$,then $\alpha = a \cdot n = (2 \hat{i} + \hat{k}) \cdot (3 \hat{j} + 2 \hat{k}) = (2)(0) + (0)(3) + (1)(2) = 2$.
Thus,$\alpha = 2$.

(D) The scalar triple product of three vectors $a, b, c$ is defined as $a \cdot(b \times c)$.
It follows the cyclic property: $a \cdot(b \times c) = b \cdot(c \times a) = c \cdot(a \times b)$.
Given the expression $w \cdot(u \times v)$,by the cyclic property,this is equal to $u \cdot(v \times w)$ and $v \cdot(w \times u)$.
Also,since the dot product is commutative,$w \cdot(u \times v) = (u \times v) \cdot w$.
However,$v \cdot(u \times w) = -(v \cdot(w \times u)) = -(w \cdot(u \times v))$.
Therefore,$v \cdot(u \times w)$ is not equal to $w \cdot(u \times v)$.

(B) The vectors representing the co-terminus edges are given by:
$AB = (5-3)\hat{i} + (-2-7)\hat{j} + (-3-4)\hat{k} = 2\hat{i} - 9\hat{j} - 7\hat{k}$
$AC = (-4-3)\hat{i} + (5-7)\hat{j} + (6-4)\hat{k} = -7\hat{i} - 2\hat{j} + 2\hat{k}$
$AD = (1-3)\hat{i} + (2-7)\hat{j} + (3-4)\hat{k} = -2\hat{i} - 5\hat{j} - 1\hat{k}$
The volume of the parallelepiped is given by the scalar triple product $|[AB, AC, AD]| = |\vec{AB} \cdot (\vec{AC} \times \vec{AD})|$.
This is equal to the absolute value of the determinant:
$|\begin{vmatrix} 2 & -9 & -7 \\ -7 & -2 & 2 \\ -2 & -5 & -1 \end{vmatrix}|$
$= |2(2 - (-10)) - (-9)(7 - (-4)) + (-7)(35 - 4)|$
$= |2(12) + 9(11) - 7(31)|$
$= |24 + 99 - 217|$
$= |123 - 217| = |-94| = 94$ cubic units.

(C) Given that $\vec{p}, \vec{q}, \vec{r}$ are non-zero,non-coplanar vectors:
The scalar triple product $[\vec{p}+\vec{q}-\vec{r}, \vec{p}-\vec{q}, \vec{q}-\vec{r}]$ can be expressed as the determinant of the coefficients of $\vec{p}, \vec{q}, \vec{r}$ multiplied by the scalar triple product $[\vec{p} \quad \vec{q} \quad \vec{r}]$.
$[\vec{p}+\vec{q}-\vec{r}, \vec{p}-\vec{q}, \vec{q}-\vec{r}] = \begin{vmatrix} 1 & 1 & -1 \\ 1 & -1 & 0 \\ 0 & 1 & -1 \end{vmatrix} [\vec{p} \quad \vec{q} \quad \vec{r}]$
Calculating the determinant:
$= [1((-1)(-1) - (0)(1)) - 1((1)(-1) - (0)(0)) + (-1)((1)(1) - (-1)(0))] [\vec{p} \quad \vec{q} \quad \vec{r}]$
$= [1(1) - 1(-1) - 1(1)] [\vec{p} \quad \vec{q} \quad \vec{r}]$
$= (1 + 1 - 1) [\vec{p} \quad \vec{q} \quad \vec{r}]$
$= [\vec{p} \quad \vec{q} \quad \vec{r}]$

(C) The volume of a parallelepiped with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$.
Given edges are $\vec{a}+\vec{b}, \vec{b}+\vec{c}, \text{ and } \vec{c}+\vec{a}$.
The volume $V = [(\vec{a}+\vec{b}) (\vec{b}+\vec{c}) (\vec{c}+\vec{a})]$.
Using the property of scalar triple product,$[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}] = (\vec{a}+\vec{b}) \cdot ((\vec{b}+\vec{c}) \times (\vec{c}+\vec{a}))$.
Expanding the cross product: $(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a}) = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a}$.
Since $\vec{c} \times \vec{c} = 0$,this simplifies to $\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a}$.
Now,$V = (\vec{a}+\vec{b}) \cdot (\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a})$.
$V = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{c} \times \vec{a}) + \vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{b} \times \vec{a}) + \vec{b} \cdot (\vec{c} \times \vec{a})$.
Terms like $\vec{a} \cdot (\vec{b} \times \vec{a})$ are zero because $\vec{a}$ is perpendicular to $\vec{b} \times \vec{a}$.
Only $\vec{a} \cdot (\vec{b} \times \vec{c})$ and $\vec{b} \cdot (\vec{c} \times \vec{a})$ remain.
$V = [\vec{a} \vec{b} \vec{c}] + [\vec{b} \vec{c} \vec{a}] = [\vec{a} \vec{b} \vec{c}] + [\vec{a} \vec{b} \vec{c}] = 2[\vec{a} \vec{b} \vec{c}]$.