MHT CET 2024 Mathematics Question Paper with Answer and Solution

(D) The probability distribution is given as:
$P(X=1)=0.15, P(X=2)=0.23, P(X=3)=0.10, P(X=4)=0.12, P(X=5)=0.20, P(X=6)=0.08, P(X=7)=0.07, P(X=8)=0.05$.
Event $E = \{ X \text{ is a prime number} \} = \{ 2, 3, 5, 7 \}$.
$P(E) = P(X=2) + P(X=3) + P(X=5) + P(X=7) = 0.23 + 0.10 + 0.20 + 0.07 = 0.60$.
Event $F = \{ X < 4 \} = \{ 1, 2, 3 \}$.
$P(F) = P(X=1) + P(X=2) + P(X=3) = 0.15 + 0.23 + 0.10 = 0.48$.
Event $E \cap F = \{ X \text{ is a prime number and } X < 4 \} = \{ 2, 3 \}$.
$P(E \cap F) = P(X=2) + P(X=3) = 0.23 + 0.10 = 0.33$.
Using the addition rule for probability:
$P(E \cup F) = P(E) + P(F) - P(E \cap F)$
$P(E \cup F) = 0.60 + 0.48 - 0.33 = 0.75$.

(A) We know that the sum of all probabilities in a probability distribution is $1$,i.e.,$\sum P(X=x) = 1$.
Given the distribution:
$k^2 + 2k + k + 2k + 5k^2 = 1$
$6k^2 + 5k - 1 = 0$
$(6k - 1)(k + 1) = 0$
Since $P(X) \geq 0$,we must have $k > 0$,so $k = \frac{1}{6}$.
Now,we need to find $P(X \geq 2)$:
$P(X \geq 2) = P(X=2) + P(X=3) + P(X=4) + P(X=5)$
$P(X \geq 2) = 2k + k + 2k + 5k^2 = 5k + 5k^2$
Substituting $k = \frac{1}{6}$:
$P(X \geq 2) = 5(\frac{1}{6}) + 5(\frac{1}{6})^2$
$P(X \geq 2) = \frac{5}{6} + \frac{5}{36} = \frac{30 + 5}{36} = \frac{35}{36}$.

(C) The sum of all probabilities in a probability distribution is always $1$.
$\therefore k + 0.3 + 0.15 + 0.15 + 0.1 + 2k = 1$
$3k + 0.7 = 1$
$3k = 0.3$
$k = 0.1$
The expected value $E(X)$ is given by $\sum x_i \cdot P(x_i)$:
$E(X) = 0(k) + 1(0.3) + 2(0.15) + 3(0.15) + 4(0.1) + 5(2k)$
$E(X) = 0(0.1) + 0.3 + 0.3 + 0.45 + 0.4 + 5(0.2)$
$E(X) = 0 + 0.3 + 0.3 + 0.45 + 0.4 + 1.0$
$E(X) = 2.45$

(B) Let $X$ denote the number of heads. When three coins are tossed,the sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$,so the total number of outcomes is $8$.
The possible values of $X$ are $0, 1, 2,$ and $3$.
$P(X=0) = P(\{TTT\}) = \frac{1}{8}$
$P(X=1) = P(\{HTT, THT, TTH\}) = \frac{3}{8}$
$P(X=2) = P(\{HHT, HTH, THH\}) = \frac{3}{8}$
$P(X=3) = P(\{HHH\}) = \frac{1}{8}$
Thus,the probability distribution is:

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

Therefore,Option $(B)$ is the correct answer.

(D) Let $P(X=0)=p_0, P(X=1)=p_1, P(X=2)=p_2, P(X=3)=p_3$.
Given $p_2 = 0.3$ and $p_3 = 2p_1$.
The sum of probabilities is $p_0 + p_1 + p_2 + p_3 = 1$.
Substituting the values,$p_0 + p_1 + 0.3 + 2p_1 = 1 \Rightarrow p_0 + 3p_1 = 0.7 \dots (i)$.
The mean is given by $E(X) = \sum x_i p_i = 0(p_0) + 1(p_1) + 2(p_2) + 3(p_3) = 1.3$.
Substituting the values,$0 + p_1 + 2(0.3) + 3(2p_1) = 1.3$.
$p_1 + 0.6 + 6p_1 = 1.3 \Rightarrow 7p_1 = 0.7 \Rightarrow p_1 = 0.1$.
From equation $(i)$,$p_0 + 3(0.1) = 0.7 \Rightarrow p_0 + 0.3 = 0.7 \Rightarrow p_0 = 0.4$.
Thus,$P(X=0) = 0.4$.

(A) Since the sum of all probabilities in a probability distribution is $1$,we have:
$\sum_{x=0}^7 P(X=x) = 1$
$0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$
$10k^2 + 9k - 1 = 0$
$(10k - 1)(k + 1) = 0$
Since $k \geq 0$,we have $k = \frac{1}{10}$.
Now,we need to find $P(X \geq 6)$:
$P(X \geq 6) = P(X=6) + P(X=7)$
$P(X \geq 6) = 2k^2 + (7k^2 + k) = 9k^2 + k$
Substituting $k = \frac{1}{10}$:
$P(X \geq 6) = 9\left(\frac{1}{10}\right)^2 + \frac{1}{10} = \frac{9}{100} + \frac{10}{100} = \frac{19}{100}$.

(B) To find $E(X^2)$,we first determine the probability mass function $P(X=x)$ from the cumulative distribution function $F(x)$ using the relation $P(X=x_i) = F(x_i) - F(x_{i-1})$.
$P(X=-1) = F(-1) = 0.3$
$P(X=0) = F(0) - F(-1) = 0.7 - 0.3 = 0.4$
$P(X=1) = F(1) - F(0) = 0.8 - 0.7 = 0.1$
$P(X=2) = F(2) - F(1) = 1 - 0.8 = 0.2$
Now,we calculate the expected value $E(X^2)$ using the formula $E(X^2) = \sum x_i^2 \cdot P(X=x_i)$:
$E(X^2) = (-1)^2(0.3) + (0)^2(0.4) + (1)^2(0.1) + (2)^2(0.2)$
$E(X^2) = (1)(0.3) + (0)(0.4) + (1)(0.1) + (4)(0.2)$
$E(X^2) = 0.3 + 0 + 0.1 + 0.8$
$E(X^2) = 1.2$

(C) We know that for a probability distribution,the sum of all probabilities must be $1$,i.e.,$\sum_{x=0}^{\infty} P(X=x) = 1$.
Substituting the given expression: $k \sum_{x=0}^{\infty} (x+1) (\frac{1}{5})^x = 1$.
Expanding the series: $k [ 1 + 2(\frac{1}{5}) + 3(\frac{1}{5})^2 + 4(\frac{1}{5})^3 + \ldots ] = 1$.
This is an arithmetico-geometric series of the form $\sum_{n=0}^{\infty} (a+nd)r^n = \frac{a}{1-r} + \frac{dr}{(1-r)^2}$ where $a=1, d=1, r=\frac{1}{5}$.
Calculating the sum: $k [ \frac{1}{1 - 1/5} + \frac{1 \times 1/5}{(1 - 1/5)^2} ] = 1$.
$k [ \frac{5}{4} + \frac{1/5}{16/25} ] = k [ \frac{5}{4} + \frac{5}{16} ] = 1$.
$k [ \frac{20+5}{16} ] = 1 \Rightarrow \frac{25k}{16} = 1$.
Therefore,$k = \frac{16}{25}$.

(C) The centroid of a tetrahedron with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,$(x_3, y_3, z_3)$,and $(x_4, y_4, z_4)$ is given by $\left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$.
Given vertices are $P(5, -7, 0)$,$Q(a, 5, 3)$,$R(4, -6, b)$,and $S(6, c, 2)$.
The centroid is $\left(\frac{5+a+4+6}{4}, \frac{-7+5-6+c}{4}, \frac{0+3+b+2}{4}\right) = \left(\frac{15+a}{4}, \frac{-8+c}{4}, \frac{b+5}{4}\right)$.
Equating this to the given centroid $(4, -3, 2)$:
$\frac{15+a}{4} = 4$ $\Rightarrow 15+a = 16$ $\Rightarrow a = 1$.
$\frac{-8+c}{4} = -3$ $\Rightarrow -8+c = -12$ $\Rightarrow c = -4$.
$\frac{b+5}{4} = 2$ $\Rightarrow b+5 = 8$ $\Rightarrow b = 3$.
Therefore,$2a + 3b + c = 2(1) + 3(3) + (-4) = 2 + 9 - 4 = 7$.

(B) Given $f(x) = \sin(2 \pi x)$.
For Rolle's theorem,$f'(C) = 0$ for some $C \in (-1, 1)$.
$f'(x) = 2 \pi \cos(2 \pi x)$.
Setting $f'(C) = 0$ gives $2 \pi \cos(2 \pi C) = 0$,which implies $\cos(2 \pi C) = 0$.
Since $C \in (-1, 1)$,we have $2 \pi C \in (-2 \pi, 2 \pi)$.
The values of $2 \pi C$ in the interval $(-2 \pi, 2 \pi)$ where $\cos(2 \pi C) = 0$ are $\frac{-3 \pi}{2}, \frac{-\pi}{2}, \frac{\pi}{2}, \frac{3 \pi}{2}$.
Dividing by $2 \pi$,we get $C = \frac{-3}{4}, \frac{-1}{4}, \frac{1}{4}, \frac{3}{4}$.
Thus,there are $4$ such values of $C$.

(A) Given that $x, y, z$ are in $A.P.$,we have $2y = x + z$ . . . $(i)$
Also,$\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are in $A.P.$,so $2 \tan^{-1} y = \tan^{-1} x + \tan^{-1} z$.
Using the formula $\tan^{-1} a + \tan^{-1} b = \tan^{-1} (\frac{a+b}{1-ab})$,we get $\tan^{-1} (\frac{2y}{1-y^2}) = \tan^{-1} (\frac{x+z}{1-xz})$.
This implies $\frac{2y}{1-y^2} = \frac{x+z}{1-xz}$.
Substituting $x+z = 2y$ from $(i)$,we get $\frac{2y}{1-y^2} = \frac{2y}{1-xz}$.
Since $y < 1$,$y$ cannot be $0$ for the progression to be non-trivial,thus $1-y^2 = 1-xz$,which simplifies to $y^2 = xz$.
Since $x, y, z$ are in both $A.P.$ and $G.P.$,it must be that $x = y = z$.

(C) The expected value $E(X)$ is calculated as follows:
$E(X) = \sum x_i P(x_i)$
$E(X) = (-2)(0.1) + (-1)(0.2) + (0)(0.2) + (1)(0.3) + (2)(0.15) + (3)(0.05)$
$E(X) = -0.2 - 0.2 + 0 + 0.3 + 0.3 + 0.15 = 0.35$
The expected value of the square $E(X^2)$ is:
$E(X^2) = \sum x_i^2 P(x_i)$
$E(X^2) = (-2)^2(0.1) + (-1)^2(0.2) + (0)^2(0.2) + (1)^2(0.3) + (2)^2(0.15) + (3)^2(0.05)$
$E(X^2) = (4)(0.1) + (1)(0.2) + 0 + (1)(0.3) + (4)(0.15) + (9)(0.05)$
$E(X^2) = 0.4 + 0.2 + 0 + 0.3 + 0.6 + 0.45 = 1.95$
The variance $Var(X)$ is given by:
$Var(X) = E(X^2) - [E(X)]^2$
$Var(X) = 1.95 - (0.35)^2$
$Var(X) = 1.95 - 0.1225 = 1.8275$

(D) The number of green balls drawn follows a binomial distribution $B(n, p)$.
Here,the total number of balls is $15 + 10 = 25$.
The probability of drawing a green ball in a single trial is $p = \frac{15}{25} = \frac{3}{5}$.
The probability of drawing a yellow ball is $q = 1 - p = \frac{10}{25} = \frac{2}{5}$.
The number of trials is $n = 10$.
The variance of a binomial distribution is given by $\sigma^2 = npq$.
Substituting the values,we get $\sigma^2 = 10 \times \frac{3}{5} \times \frac{2}{5} = 10 \times \frac{6}{25} = \frac{60}{25} = \frac{12}{5}$.

(B) Let the vertices be $P(0,3,0)$,$Q(0,0,4)$,and $R(0,3,4)$.
The side lengths are:
$PQ = \sqrt{(0-0)^2 + (0-3)^2 + (4-0)^2} = \sqrt{0+9+16} = 5$
$QR = \sqrt{(0-0)^2 + (3-0)^2 + (4-4)^2} = \sqrt{0+9+0} = 3$
$PR = \sqrt{(0-0)^2 + (3-3)^2 + (4-0)^2} = \sqrt{0+0+16} = 4$
The incentre $I(x,y,z)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ and opposite side lengths $a, b, c$ is given by:
$I = \left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}, \frac{az_1 + bz_2 + cz_3}{a+b+c} \right)$
Here,$a=QR=3$,$b=PR=4$,$c=PQ=5$.
$I = \left( \frac{3(0)+4(0)+5(0)}{3+4+5}, \frac{3(3)+4(0)+5(3)}{3+4+5}, \frac{3(0)+4(4)+5(4)}{3+4+5} \right)$
$I = \left( \frac{0}{12}, \frac{9+0+15}{12}, \frac{0+16+20}{12} \right)$
$I = \left( 0, \frac{24}{12}, \frac{36}{12} \right) = (0, 2, 3)$

(C) Let the vertices be $A(0,2,1)$,$B(-2,0,0)$,and $C(-2,0,2)$.
Calculate the side lengths:
$a = BC = \sqrt{(-2 - (-2))^2 + (0 - 0)^2 + (2 - 0)^2} = \sqrt{0 + 0 + 4} = 2$
$b = AC = \sqrt{(-2 - 0)^2 + (0 - 2)^2 + (2 - 1)^2} = \sqrt{4 + 4 + 1} = 3$
$c = AB = \sqrt{(-2 - 0)^2 + (0 - 2)^2 + (0 - 1)^2} = \sqrt{4 + 4 + 1} = 3$
The incenter $I$ is given by $\frac{aA + bB + cC}{a + b + c}$.
$I = \frac{2(0,2,1) + 3(-2,0,0) + 3(-2,0,2)}{2 + 3 + 3}$
$I = \frac{(0,4,2) + (-6,0,0) + (-6,0,6)}{8}$
$I = \frac{(-12, 4, 8)}{8} = \left(-\frac{12}{8}, \frac{4}{8}, \frac{8}{8}\right) = \left(-\frac{3}{2}, \frac{1}{2}, 1\right)$.

(B) Let the vertices be $A(1,2,0)$,$B(1,0,2)$,and $C(0,3,1)$.
The vectors representing the sides are $\vec{AB} = (1-1)\hat{i} + (0-2)\hat{j} + (2-0)\hat{k} = -2\hat{j} + 2\hat{k}$ and $\vec{AC} = (0-1)\hat{i} + (3-2)\hat{j} + (1-0)\hat{k} = -\hat{i} + \hat{j} + \hat{k}$.
The area of $\triangle ABC$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$.
Calculating the cross product: $\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -2 & 2 \\ -1 & 1 & 1 \end{vmatrix} = \hat{i}(-2-2) - \hat{j}(0 - (-2)) + \hat{k}(0-2) = -4\hat{i} - 2\hat{j} - 2\hat{k}$.
The magnitude is $|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + (-2)^2 + (-2)^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6}$.
Therefore,the area is $\frac{1}{2} \times 2\sqrt{6} = \sqrt{6}$ sq. units.

(A) Let the position vectors of the two points be $\bar{b} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\bar{c} = -\hat{i} + 4\hat{j} + \hat{k}$.
The direction vector of the line joining these points is $\bar{v} = \bar{c} - \bar{b} = (-\hat{i} - \hat{i}) + (4\hat{j} - 2\hat{j}) + (\hat{k} - 2\hat{k}) = -2\hat{i} + 2\hat{j} - \hat{k}$.
The equation of a line passing through point $\bar{a} = 2\hat{i} - \hat{j} + \hat{k}$ and parallel to vector $\bar{v}$ is given by $\bar{r} = \bar{a} + \lambda\bar{v}$.
Substituting the values,we get $\bar{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(-2\hat{i} + 2\hat{j} - \hat{k})$.

(C) Let the angles made by the line with the positive $X$,$Y$,and $Z$ axes be $\alpha$,$\beta$,and $\gamma$ respectively.
Given $\alpha = 45^{\circ}$ and $\beta = \gamma$.
Using the direction cosine property $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$:
$\cos^2 45^{\circ} + \cos^2 \beta + \cos^2 \beta = 1$
$\left(\frac{1}{\sqrt{2}}\right)^2 + 2\cos^2 \beta = 1$
$\frac{1}{2} + 2\cos^2 \beta = 1$
$2\cos^2 \beta = \frac{1}{2}$
$\cos^2 \beta = \frac{1}{4}$
$\cos \beta = \frac{1}{2}$ (since angles are acute)
$\beta = 60^{\circ}$.
Thus,$\beta = 60^{\circ}$ and $\gamma = 60^{\circ}$.
The sum of the angles is $\alpha + \beta + \gamma = 45^{\circ} + 60^{\circ} + 60^{\circ} = 165^{\circ}$.

(C) Since the direction cosines of the line $PQ$ are equal and positive,let them be $l, m, n$. Since $l=m=n$ and $l^2+m^2+n^2=1$,we have $3l^2=1$,so $l=m=n=\frac{1}{\sqrt{3}}$.
The direction ratios of the line $PQ$ are proportional to $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$,which can be taken as $1, 1, 1$.
The equation of the line $PQ$ passing through $P(2,-1,2)$ is $\frac{x-2}{1} = \frac{y+1}{1} = \frac{z-2}{1} = k$.
Thus,any point on the line is $(k+2, k-1, k+2)$.
Since this point $Q$ lies on the plane $2x+y+z=9$,we have $2(k+2) + (k-1) + (k+2) = 9$.
$4k + 5 = 9$ $\Rightarrow 4k = 4$ $\Rightarrow k = 1$.
Substituting $k=1$,the coordinates of $Q$ are $(3, 0, 3)$.
The length $PQ = \sqrt{(3-2)^2 + (0-(-1))^2 + (3-2)^2} = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ units.

(A) Given the curve equation: $y = a\left(e^{\frac{x}{a}} + e^{-\frac{x}{a}}\right)$.
To find the slope of the tangent,we differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = a \left( e^{\frac{x}{a}} \cdot \frac{1}{a} + e^{-\frac{x}{a}} \cdot (-\frac{1}{a}) \right) = e^{\frac{x}{a}} - e^{-\frac{x}{a}}$.
Since the tangent is parallel to the $X$-axis,its slope must be zero:
$\frac{dy}{dx} = 0 \implies e^{\frac{x}{a}} - e^{-\frac{x}{a}} = 0$.
$e^{\frac{x}{a}} = e^{-\frac{x}{a}}$.
Taking the natural logarithm on both sides:
$\frac{x}{a} = -\frac{x}{a} \implies \frac{2x}{a} = 0 \implies x = 0$.
Thus,the abscissa of the point is $0$.

(B) Given: $|\overline{a}|=2, |\overline{b}|=4, |\overline{c}|=4$.
According to the condition,the projection of $\overline{b}$ on $\overline{a}$ is equal to the projection of $\overline{c}$ on $\overline{a}$.
$\frac{\overline{b} \cdot \overline{a}}{|\overline{a}|} = \frac{\overline{c} \cdot \overline{a}}{|\overline{a}|}$
$\implies \overline{b} \cdot \overline{a} = \overline{c} \cdot \overline{a}$
$\implies (\overline{b} - \overline{c}) \cdot \overline{a} = 0$ ... $(i)$
Also,$\overline{b}$ is perpendicular to $\overline{c}$,so $\overline{b} \cdot \overline{c} = 0$.
Now,consider $|\overline{a} + \overline{b} - \overline{c}|^2 = |\overline{a}|^2 + |\overline{b} - \overline{c}|^2 + 2\overline{a} \cdot (\overline{b} - \overline{c})$.
From $(i)$,$\overline{a} \cdot (\overline{b} - \overline{c}) = 0$.
So,$|\overline{a} + \overline{b} - \overline{c}|^2 = |\overline{a}|^2 + |\overline{b}|^2 + |\overline{c}|^2 - 2(\overline{b} \cdot \overline{c})$.
Since $\overline{b} \cdot \overline{c} = 0$,we have:
$|\overline{a} + \overline{b} - \overline{c}|^2 = (2)^2 + (4)^2 + (4)^2 - 0 = 4 + 16 + 16 = 36$.
Therefore,$|\overline{a} + \overline{b} - \overline{c}| = \sqrt{36} = 6$.

(C) The projection of $\overline{a}$ on $\overline{c}$ is given by $\frac{\overline{a} \cdot \overline{c}}{|\overline{c}|} = \frac{10}{3}$.
$|\overline{c}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1+4+4} = 3$.
So,$\frac{(\alpha \hat{i} + 3 \hat{j} - \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{3} = \frac{10}{3}$.
$\alpha + 6 + 2 = 10 \Rightarrow \alpha = 2$.
Now,$\overline{b} \times \overline{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -\beta & 4 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(2\beta - 8) - \hat{j}(-6 - 4) + \hat{k}(6 + \beta) = (2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$.
Given $\overline{b} \times \overline{c} = -6\hat{i} + 10\hat{j} + 7\hat{k}$,comparing components:
$6 + \beta = 7 \Rightarrow \beta = 1$.
Thus,$\alpha^2 + \beta^2 - \alpha\beta = 2^2 + 1^2 - (2)(1) = 4 + 1 - 2 = 3$.

(B) First,we find the vectors $\overline{AB}$ and $\overline{CD}$:
$\overline{AB} = (1-2)\hat{i} + (-4 - (-3))\hat{j} + (-2-0)\hat{k} = -\hat{i} - \hat{j} - 2\hat{k}$
$\overline{CD} = (7-4)\hat{i} + (0-6)\hat{j} + (10-8)\hat{k} = 3\hat{i} - 6\hat{j} + 2\hat{k}$
The scalar projection of vector $\overline{AB}$ on vector $\overline{CD}$ is given by the formula:
$\text{Projection} = \frac{\overline{AB} \cdot \overline{CD}}{|\overline{CD}|}$
Calculate the dot product $\overline{AB} \cdot \overline{CD}$:
$\overline{AB} \cdot \overline{CD} = (-1)(3) + (-1)(-6) + (-2)(2) = -3 + 6 - 4 = -1$
Calculate the magnitude of $\overline{CD}$:
$|\overline{CD}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$
Thus,the scalar projection is:
$\frac{-1}{7}$
Note: The magnitude of the projection is $|\frac{-1}{7}| = \frac{1}{7}$.

(D) Let the direction ratios of the required line be $a, b, c$.
Since the line is perpendicular to the lines with direction ratios $(1, 2, 3)$ and $(-3, 2, 5)$,we have the following equations:
$a + 2b + 3c = 0$ $(i)$
$-3a + 2b + 5c = 0$ (ii)
Using the cross-multiplication method to solve for $a, b, c$:
$\frac{a}{(2)(5) - (3)(2)} = \frac{b}{(3)(-3) - (1)(5)} = \frac{c}{(1)(2) - (2)(-3)}$
$\frac{a}{10 - 6} = \frac{b}{-9 - 5} = \frac{c}{2 + 6}$
$\frac{a}{4} = \frac{b}{-14} = \frac{c}{8}$
Dividing by $2$,we get the direction ratios as $(2, -7, 4)$.
The equation of the line passing through $(-1, 3, -2)$ with direction ratios $(2, -7, 4)$ is:
$\frac{x - (-1)}{2} = \frac{y - 3}{-7} = \frac{z - (-2)}{4}$
$\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}$

(D) Let the position vector of point $A$ be $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$.
Let the two given vectors be $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = \hat{i} + 3 \hat{j} + 2 \hat{k}$.
The direction of the line is perpendicular to both $\vec{b}$ and $\vec{c}$,so it is parallel to $\vec{b} \times \vec{c}$.
$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(2 - (-3)) - \hat{j}(4 - (-1)) + \hat{k}(6 - 1) = 5 \hat{i} - 5 \hat{j} + 5 \hat{k}$.
We can take the direction vector as $\vec{v} = \hat{i} - \hat{j} + \hat{k}$ (dividing by $5$).
The equation of the line is $\vec{r} = \vec{a} + \lambda \vec{v}$,which is $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$.

(A) The vector equations of the given lines are $\bar{r}=\hat{j}+\lambda(3 \hat{i}-2 \hat{j}+\hat{k})$ and $\bar{r}=-4 \hat{i}+3 \hat{j}-2 \hat{k}+\mu(3 \hat{i}-2 \hat{j}+\hat{k})$.
The distance between the parallel lines $\bar{r}=\bar{a}_1+\lambda \bar{b}$ and $\bar{r}=\bar{a}_2+\mu \bar{b}$ is given by $d=\left|\frac{(\bar{a}_2-\bar{a}_1) \times \bar{b}}{|\bar{b}|}\right|$.
Here,$\bar{a}_1=\hat{j}$,$\bar{a}_2=-4 \hat{i}+3 \hat{j}-2 \hat{k}$,and $\bar{b}=3 \hat{i}-2 \hat{j}+\hat{k}$.
Therefore,$\bar{a}_2-\bar{a}_1=-4 \hat{i}+2 \hat{j}-2 \hat{k}$.
The cross product $(\bar{a}_2-\bar{a}_1) \times \bar{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 2 & -2 \\ 3 & -2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(-4+6) + \hat{k}(8-6) = -2 \hat{i}-2 \hat{j}+2 \hat{k}$.
The magnitude $|\bar{b}| = \sqrt{3^2+(-2)^2+1^2} = \sqrt{9+4+1} = \sqrt{14}$.
Thus,$d = \frac{|-2 \hat{i}-2 \hat{j}+2 \hat{k}|}{\sqrt{14}} = \frac{\sqrt{(-2)^2+(-2)^2+2^2}}{\sqrt{14}} = \sqrt{\frac{4+4+4}{14}} = \sqrt{\frac{12}{14}} = \sqrt{\frac{6}{7}}$ units.

(A) Let the point be $P(a, 6, 9)$ and its mirror image be $P'(20, b, -a-9)$.
The midpoint $M$ of $PP'$ is $\left(\frac{a+20}{2}, \frac{6+b}{2}, \frac{9-a-9}{2}\right) = \left(\frac{a+20}{2}, \frac{6+b}{2}, -\frac{a}{2}\right)$.
Since $M$ lies on the line $\frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9} = k$,we have:
$\frac{\frac{a+20}{2}-3}{7} = \frac{\frac{6+b}{2}-2}{5} = \frac{-\frac{a}{2}-1}{-9} = k$.
From the first and third parts: $\frac{a+14}{14} = \frac{a+2}{18} \implies 18a + 252 = 14a + 28 \implies 4a = -224 \implies a = -56$.
Substituting $a = -56$ into the line equation: $\frac{-56+14}{14} = \frac{6+b-4}{10} \implies -3 = \frac{b+2}{10} \implies b+2 = -30 \implies b = -32$.
Thus,$|a+b| = |-56 - 32| = |-88| = 88$.

(B) The direction vectors of the lines $L_1$ and $L_2$ are $\vec{v_1} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{v_2} = \hat{i} + 2\hat{j} + 3\hat{k}$ respectively.
$A$ vector perpendicular to both $L_1$ and $L_2$ is given by the cross product $\vec{n} = \vec{v_1} \times \vec{v_2}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-4) - \hat{j}(9-2) + \hat{k}(6-1) = -\hat{i} - 7\hat{j} + 5\hat{k}$.
The magnitude of this vector is $|\vec{n}| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$.
The required unit vector is $\frac{\vec{n}}{|\vec{n}|} = \frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$.

(A) The given Cartesian equation is $2x - 2 = 3y + 1 = 6z - 2$.
Divide the entire equation by the least common multiple of the coefficients of $x, y, z$,which is $6$:
$\frac{2(x - 1)}{6} = \frac{3(y + 1/3)}{6} = \frac{6(z - 1/3)}{6}$
This simplifies to:
$\frac{x - 1}{3} = \frac{y + 1/3}{2} = \frac{z - 1/3}{1}$
The line passes through the point $(1, -1/3, 1/3)$ and has direction ratios $(3, 2, 1)$.
The vector equation of a line passing through point $\vec{a}$ with direction vector $\vec{b}$ is $\vec{r} = \vec{a} + \lambda\vec{b}$.
Substituting the values,we get:
$\vec{r} = \left(\hat{i} - \frac{1}{3}\hat{j} + \frac{1}{3}\hat{k}\right) + \lambda(3\hat{i} + 2\hat{j} + \hat{k})$.

(C) Let $\frac{x-1}{2} = \frac{y+3}{-1} = \frac{z+1}{-2} = \lambda$.
Any general point $Q$ on the line is given by $Q(2\lambda+1, -\lambda-3, -2\lambda-1)$.
The vector $\vec{AQ}$ is $(2\lambda+1-1, -\lambda-3-(-2), -2\lambda-1-(-3)) = (2\lambda, -\lambda-1, -2\lambda+2)$.
Since $AQ$ is perpendicular to the line with direction ratios $(2, -1, -2)$, their dot product must be zero:
$2(2\lambda) - 1(-\lambda-1) - 2(-2\lambda+2) = 0$.
$4\lambda + \lambda + 1 + 4\lambda - 4 = 0$.
$9\lambda - 3 = 0 \Rightarrow \lambda = \frac{1}{3}$.
Substituting $\lambda = \frac{1}{3}$ into $Q$, we get $Q = (\frac{5}{3}, -\frac{10}{3}, -\frac{5}{3})$.
The length $AQ = \sqrt{(\frac{5}{3}-1)^2 + (-\frac{10}{3}-(-2))^2 + (-\frac{5}{3}-(-3))^2}$.
$AQ = \sqrt{(\frac{2}{3})^2 + (-\frac{4}{3})^2 + (\frac{4}{3})^2} = \sqrt{\frac{4}{9} + \frac{16}{9} + \frac{16}{9}} = \sqrt{\frac{36}{9}} = \sqrt{4} = 2 \text{ units}$.

(A) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ intersect if and only if the determinant of the matrix formed by their direction ratios and the difference of their points is zero:
$\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$
Given the lines:
Line $1$: $(x_1, y_1, z_1) = (1, 2, 1)$ and $(a_1, b_1, c_1) = (2, 3, 4)$
Line $2$: $(x_2, y_2, z_2) = (3, k, 0)$ and $(a_2, b_2, c_2) = (-1, 2, 1)$
Substituting these values into the determinant:
$\left|\begin{array}{ccc} 3-1 & k-2 & 0-1 \\ 2 & 3 & 4 \\ -1 & 2 & 1 \end{array}\right| = 0$
$\left|\begin{array}{ccc} 2 & k-2 & -1 \\ 2 & 3 & 4 \\ -1 & 2 & 1 \end{array}\right| = 0$
Expanding along the first row:
$2(3(1) - 4(2)) - (k-2)(2(1) - 4(-1)) - 1(2(2) - 3(-1)) = 0$
$2(3 - 8) - (k-2)(2 + 4) - 1(4 + 3) = 0$
$2(-5) - (k-2)(6) - 1(7) = 0$
$-10 - 6k + 12 - 7 = 0$
$-6k - 5 = 0$
$-6k = 5$
$k = \frac{-5}{6}$

(C) The required line passes through the point $(3, 1, 2)$ and is perpendicular to the lines with direction vectors $\vec{v_1} = (1, 2, 3)$ and $\vec{v_2} = (-3, 2, 5)$.
The direction vector $\vec{b}$ of the required line is given by the cross product of $\vec{v_1}$ and $\vec{v_2}$:
$\vec{b} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{vmatrix}$
$\vec{b} = \hat{i}(2 \times 5 - 3 \times 2) - \hat{j}(1 \times 5 - 3 \times (-3)) + \hat{k}(1 \times 2 - 2 \times (-3))$
$\vec{b} = \hat{i}(10 - 6) - \hat{j}(5 + 9) + \hat{k}(2 + 6) = 4\hat{i} - 14\hat{j} + 8\hat{k}$
We can simplify the direction vector by dividing by $2$: $\vec{b'} = 2\hat{i} - 7\hat{j} + 4\hat{k}$.
The equation of the line passing through $(x_1, y_1, z_1)$ with direction vector $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Substituting the point $(3, 1, 2)$ and direction vector $(2, -7, 4)$,we get:
$\frac{x-3}{2} = \frac{y-1}{-7} = \frac{z-2}{4}$.

(B) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ intersect if and only if the determinant of the vector formed by the difference of points and the direction vectors is zero:
$\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$
Given points are $(-1, -k, 4)$ and $(-10, -1, 1)$,and direction vectors are $(-10, -1, 1)$ and $(-1, -3, 4)$.
Substituting the values:
$\left|\begin{array}{ccc} -10-(-1) & -1-(-k) & 1-4 \\ -10 & -1 & 1 \\ -1 & -3 & 4 \end{array}\right| = 0$
$\Rightarrow \left|\begin{array}{ccc} -9 & k-1 & -3 \\ -10 & -1 & 1 \\ -1 & -3 & 4 \end{array}\right| = 0$
Expanding the determinant along the first row:
$-9((-1)(4) - (1)(-3)) - (k-1)((-10)(4) - (1)(-1)) - 3((-10)(-3) - (-1)(-1)) = 0$
$-9(-4 + 3) - (k-1)(-40 + 1) - 3(30 - 1) = 0$
$-9(-1) - (k-1)(-39) - 3(29) = 0$
$9 + 39(k-1) - 87 = 0$
$39(k-1) = 78$
$k-1 = 2$
$k = 3$

(B) Lines $L_1$ and $L_2$ are parallel to the vectors $\vec{b}_1 = 5\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b}_2 = 4\hat{i} + 3\hat{j} + 5\hat{k}$ respectively.
The unit vector perpendicular to both $L_1$ and $L_2$ is given by $\hat{n} = \pm \frac{\vec{b}_1 \times \vec{b}_2}{|\vec{b}_1 \times \vec{b}_2|}$.
Calculating the cross product: $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 2 & 1 \\ 4 & 3 & 5 \end{vmatrix} = \hat{i}(10-3) - \hat{j}(25-4) + \hat{k}(15-8) = 7\hat{i} - 21\hat{j} + 7\hat{k}$.
The magnitude is $|\vec{b}_1 \times \vec{b}_2| = \sqrt{7^2 + (-21)^2 + 7^2} = \sqrt{49 + 441 + 49} = \sqrt{539} = 7\sqrt{11}$.
Thus,$\hat{n} = \pm \frac{7(\hat{i} - 3\hat{j} + \hat{k})}{7\sqrt{11}} = \pm \frac{\hat{i} - 3\hat{j} + \hat{k}}{\sqrt{11}}$.
Comparing with the given options,option $B$ is correct.

(A) The given equations of the lines are $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$.
The lines pass through points $P_1(1, -2, 1)$ and $P_2(3, k, 0)$ with direction ratios $\vec{v_1} = (2, 3, 4)$ and $\vec{v_2} = (1, 2, 1)$.
For two lines to intersect,the shortest distance between them must be zero,which implies the scalar triple product of the vector joining the points and the direction vectors must be zero:
$\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right|=0$
Substituting the values:
$\left|\begin{array}{ccc} 3-1 & k-(-2) & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right|=0$
$\left|\begin{array}{ccc} 2 & k+2 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{array}\right|=0$
Expanding the determinant along the first row:
$2(3(1) - 4(2)) - (k+2)(2(1) - 4(1)) - 1(2(2) - 3(1)) = 0$
$2(3-8) - (k+2)(2-4) - 1(4-3) = 0$
$2(-5) - (k+2)(-2) - 1(1) = 0$
$-10 + 2k + 4 - 1 = 0$
$2k - 7 = 0$
$k = \frac{7}{2}$

(A) Let $\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1} = \lambda$. Then $x = 3\lambda + 7, y = 1 - \lambda, z = \lambda - 2$.
Let $\frac{x}{2} = \frac{y-7}{3} = \frac{z}{1} = \mu$. Then $x = 2\mu, y = 3\mu + 7, z = \mu$.
Coordinates of point $A$ on the first line are $(3\lambda + 7, 1 - \lambda, \lambda - 2)$.
Coordinates of point $B$ on the second line are $(2\mu, 3\mu + 7, \mu)$.
The direction ratios of line $AB$ are $(3\lambda - 2\mu + 7, -\lambda - 3\mu - 6, \lambda - \mu - 2)$.
Since the direction ratios of the line are $1, -4, 2$,we have:
$\frac{3\lambda - 2\mu + 7}{1} = \frac{-\lambda - 3\mu - 6}{-4} = \frac{\lambda - \mu - 2}{2}$.
From $\frac{3\lambda - 2\mu + 7}{1} = \frac{\lambda + 3\mu + 6}{4}$,we get $12\lambda - 8\mu + 28 = \lambda + 3\mu + 6$,which simplifies to $11\lambda - 11\mu + 22 = 0$,or $\lambda - \mu + 2 = 0$ $(i)$.
From $\frac{\lambda + 3\mu + 6}{4} = \frac{\lambda - \mu - 2}{2}$,we get $\lambda + 3\mu + 6 = 2\lambda - 2\mu - 4$,which simplifies to $\lambda - 5\mu - 10 = 0$ $(ii)$.
Solving $(i)$ and $(ii)$,we subtract $(ii)$ from $(i)$: $(\lambda - \mu + 2) - (\lambda - 5\mu - 10) = 0$,so $4\mu + 12 = 0$,which gives $\mu = -3$.
Substituting $\mu = -3$ into $(i)$,$\lambda - (-3) + 2 = 0$,so $\lambda = -5$.
Thus,$A = (3(-5) + 7, 1 - (-5), -5 - 2) = (-8, 6, -7)$ and $B = (2(-3), 3(-3) + 7, -3) = (-6, -2, -3)$.

(A) Let the line be $\frac{x+3}{5}=\frac{y+1}{2}=\frac{z+4}{3}=\lambda$.
Any point $P$ on the line is given by $P(5\lambda-3, 2\lambda-1, 3\lambda-4)$.
The given point is $A(0, 2, 3)$.
The direction ratios of the line $AP$ are $(5\lambda-3-0, 2\lambda-1-2, 3\lambda-4-3)$,which simplifies to $(5\lambda-3, 2\lambda-3, 3\lambda-7)$.
Since $AP$ is perpendicular to the given line with direction ratios $(5, 2, 3)$,their dot product must be zero:
$5(5\lambda-3) + 2(2\lambda-3) + 3(3\lambda-7) = 0$.
$25\lambda - 15 + 4\lambda - 6 + 9\lambda - 21 = 0$.
$38\lambda - 42 = 0$.
$\lambda = \frac{42}{38} = \frac{21}{19}$.
Substituting $\lambda = \frac{21}{19}$ into the coordinates of $P$:
$x = 5(\frac{21}{19}) - 3 = \frac{105-57}{19} = \frac{48}{19}$.
$y = 2(\frac{21}{19}) - 1 = \frac{42-19}{19} = \frac{23}{19}$.
$z = 3(\frac{21}{19}) - 4 = \frac{63-76}{19} = \frac{-13}{19}$.
Thus,the foot of the perpendicular is $\left(\frac{48}{19}, \frac{23}{19}, \frac{-13}{19}\right)$.

(C) The equation of a plane passing through three points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by the determinant equation:
$\left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{array}\right| = 0$
Substituting the given points $(3,1,1)$,$(1,2,3)$,and $(-1,4,2)$:
$\left|\begin{array}{ccc} x-3 & y-1 & z-1 \\ 1-3 & 2-1 & 3-1 \\ -1-3 & 4-1 & 2-1 \end{array}\right| = 0$
$\Rightarrow \left|\begin{array}{ccc} x-3 & y-1 & z-1 \\ -2 & 1 & 2 \\ -4 & 3 & 1 \end{array}\right| = 0$
Expanding along the first row:
$(x-3)(1-6) - (y-1)(-2+8) + (z-1)(-6+4) = 0$
$-5(x-3) - 6(y-1) - 2(z-1) = 0$
$-5x + 15 - 6y + 6 - 2z + 2 = 0$
$-5x - 6y - 2z + 23 = 0$
Multiplying by $-1$:
$5x + 6y + 2z - 23 = 0$

(C) The normal vector to the plane $x-y+z=3$ is $\vec{n} = (1, -1, 1)$.
Let $Q = (3, 1, 7)$. The line passing through $Q$ and perpendicular to the plane is given by $\frac{x-3}{1} = \frac{y-1}{-1} = \frac{z-7}{1} = \lambda$.
Any point on this line is $(\lambda+3, -\lambda+1, \lambda+7)$.
For the intersection point $M$ with the plane,we have $(\lambda+3) - (-\lambda+1) + (\lambda+7) = 3$,which simplifies to $3\lambda + 9 = 3$,so $3\lambda = -6$,giving $\lambda = -2$.
Thus,$M = (-2+3, -(-2)+1, -2+7) = (1, 3, 5)$.
Since $M$ is the midpoint of $PQ$,if $P = (a, b, c)$,then $\frac{3+a}{2} = 1, \frac{1+b}{2} = 3, \frac{7+c}{2} = 5$.
This gives $a = -1, b = 5, c = 3$,so $P = (-1, 5, 3)$.
The plane passes through $P(-1, 5, 3)$ and contains the line $\frac{x}{1} = \frac{y}{2} = \frac{z}{1}$,which passes through the origin $O(0, 0, 0)$ with direction vector $\vec{v} = (1, 2, 1)$.
The normal to the required plane is $\vec{n'} = \vec{OP} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 5 & 3 \\ 1 & 2 & 1 \end{vmatrix} = \hat{i}(5-6) - \hat{j}(-1-3) + \hat{k}(-2-5) = -\hat{i} + 4\hat{j} - 7\hat{k}$.
The equation of the plane is $-1(x-0) + 4(y-0) - 7(z-0) = 0$,which is $-x + 4y - 7z = 0$,or $x - 4y + 7z = 0$.

(B) Let $M$ be the midpoint of the line segment $PQ$.
The coordinates of $M$ are given by $\left(\frac{1+3}{2}, \frac{2+4}{2}, \frac{5+3}{2}\right) = (2, 3, 4)$.
The direction ratios of the line $PQ$ are $(3-1, 4-2, 3-5) = (2, 2, -2)$.
Since the plane is perpendicular to $PQ$,the normal vector to the plane is $\vec{n} = 2\hat{i} + 2\hat{j} - 2\hat{k}$.
The equation of the plane passing through $(x_0, y_0, z_0)$ with normal vector $(a, b, c)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Substituting the values,we get $2(x-2) + 2(y-3) - 2(z-4) = 0$.
Dividing by $2$,we get $(x-2) + (y-3) - (z-4) = 0$.
$x + y - z - 2 - 3 + 4 = 0$.
$x + y - z - 1 = 0$.

(C) Let the required plane be $P_1$. It contains the line $L_1: \frac{x}{3}=\frac{y}{2}=\frac{z}{4}$,so its normal vector $\vec{n}_1$ is perpendicular to $\vec{v}_1 = (3, 2, 4)$.
Let $P_2$ be the plane containing $L_2: \frac{x}{4}=\frac{y}{3}=\frac{z}{2}$ and $L_3: \frac{x}{2}=\frac{y}{-4}=\frac{z}{3}$.
The normal vector $\vec{n}_2$ of $P_2$ is $\vec{v}_2 \times \vec{v}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 2 \\ 2 & -4 & 3 \end{vmatrix} = \hat{i}(9+8) - \hat{j}(12-4) + \hat{k}(-16-6) = 17\hat{i} - 8\hat{j} - 22\hat{k}$.
Since $P_1 \perp P_2$,$\vec{n}_1$ is perpendicular to $\vec{n}_2 = (17, -8, -22)$.
Thus,$\vec{n}_1 = \vec{v}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 4 \\ 17 & -8 & -22 \end{vmatrix} = \hat{i}(-44+32) - \hat{j}(-66-68) + \hat{k}(-24-34) = -12\hat{i} + 134\hat{j} - 58\hat{k}$.
Dividing by $-2$,we get the normal vector $(6, -67, 29)$.
The equation of the plane is $6x - 67y + 29z = 0$.

(C) Let the required plane be $P_1$. It contains the line $L_1: \frac{x}{2}=\frac{y}{3}=\frac{z}{4}$,so its normal vector $\vec{n}_1$ is perpendicular to $\vec{v}_1 = (2, 3, 4)$.
Let $P_2$ be the plane containing the lines $L_2: \frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $L_3: \frac{x}{4}=\frac{y}{2}=\frac{z}{3}$.
The normal vector $\vec{n}_2$ of $P_2$ is given by the cross product of the direction vectors of $L_2$ and $L_3$:
$\vec{n}_2 = (3, 4, 2) \times (4, 2, 3) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{vmatrix} = \hat{i}(12-4) - \hat{j}(9-8) + \hat{k}(6-16) = (8, -1, -10)$.
Thus,the equation of plane $P_2$ is $8x - y - 10z = 0$.
Since $P_1$ is perpendicular to $P_2$,its normal $\vec{n}_1$ is perpendicular to $\vec{n}_2 = (8, -1, -10)$.
Also,$\vec{n}_1$ is perpendicular to $\vec{v}_1 = (2, 3, 4)$.
Therefore,$\vec{n}_1 = \vec{v}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 8 & -1 & -10 \end{vmatrix} = \hat{i}(-30+4) - \hat{j}(-20-32) + \hat{k}(-2-24) = (-26, 52, -26)$.
Dividing by $-26$,we get the normal vector $(1, -2, 1)$.
The equation of the plane is $1(x-0) - 2(y-0) + 1(z-0) = 0$,which is $x - 2y + z = 0$.

(B) The required plane passes through the point $(1,1,1)$ and is perpendicular to the planes $2x-y-2z=5$ and $3x-6y+2z=7$. The normal vectors of these planes are $\vec{n_1} = 2\hat{i} - \hat{j} - 2\hat{k}$ and $\vec{n_2} = 3\hat{i} - 6\hat{j} + 2\hat{k}$. The normal vector $\vec{n}$ of the required plane is $\vec{n_1} \times \vec{n_2}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -2 \\ 3 & -6 & 2 \end{vmatrix} = \hat{i}(-2-12) - \hat{j}(4+6) + \hat{k}(-12+3) = -14\hat{i} - 10\hat{j} - 9\hat{k}$.
The equation of the plane is $a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$.
$-14(x-1) - 10(y-1) - 9(z-1) = 0$.
$-14x + 14 - 10y + 10 - 9z + 9 = 0$.
$-14x - 10y - 9z + 33 = 0$.
$14x + 10y + 9z = 33$ is incorrect in the provided options,let's recheck the determinant calculation.
$\begin{vmatrix} x-1 & y-1 & z-1 \\ 2 & -1 & -2 \\ 3 & -6 & 2 \end{vmatrix} = (x-1)(-2-12) - (y-1)(4+6) + (z-1)(-12+3) = -14(x-1) - 10(y-1) - 9(z-1) = -14x + 14 - 10y + 10 - 9z + 9 = -14x - 10y - 9z + 33 = 0$.
Thus,$14x + 10y + 9z = 33$.

(D) The equation of a plane passing through $(1, -2, 1)$ is $a(x - 1) + b(y + 2) + c(z - 1) = 0$.
Since the plane is perpendicular to the planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,its normal vector $\vec{n} = (a, b, c)$ is perpendicular to the normal vectors $\vec{n_1} = (2, -2, 1)$ and $\vec{n_2} = (1, -1, 2)$.
Thus,$\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 1 \\ 1 & -1 & 2 \end{vmatrix} = \hat{i}(-4 + 1) - \hat{j}(4 - 1) + \hat{k}(-2 + 2) = -3\hat{i} - 3\hat{j} + 0\hat{k}$.
We can take the normal vector as $(1, 1, 0)$.
The equation of the plane is $1(x - 1) + 1(y + 2) + 0(z - 1) = 0$,which simplifies to $x + y + 1 = 0$.
The distance of the plane $x + y + 0z + 1 = 0$ from the point $(1, 2, 2)$ is given by $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} = \frac{|1(1) + 1(2) + 0(2) + 1|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{|1 + 2 + 1|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$ units.

(A) The normal vector to the plane $P_1$ is given by $\overline{n_1} = \overline{a} \times \overline{b}$.
The normal vector to the plane $P_2$ is given by $\overline{n_2} = \overline{c} \times \overline{d}$.
Given the condition $(\overline{a} \times \overline{b}) \times (\overline{c} \times \overline{d}) = \overline{0}$,it implies that the vector $\overline{n_1}$ is parallel to the vector $\overline{n_2}$.
Since the normal vectors of the two planes are parallel,the planes $P_1$ and $P_2$ are parallel to each other.
The angle between two parallel planes is $0$.

(C) Let the point be $(x_1, y_1, z_1) = (-1, 2, -3)$.
The direction ratios of the two lines are $(a_1, b_1, c_1) = (3, 2, -4)$ and $(a_2, b_2, c_2) = (2, -3, 2)$.
The equation of the plane passing through $(x_1, y_1, z_1)$ and parallel to vectors $\vec{b_1} = (a_1, b_1, c_1)$ and $\vec{b_2} = (a_2, b_2, c_2)$ is given by the determinant:
$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$
Substituting the values:
$\begin{vmatrix} x + 1 & y - 2 & z + 3 \\ 3 & 2 & -4 \\ 2 & -3 & 2 \end{vmatrix} = 0$
Expanding along the first row:
$(x + 1)(2(2) - (-4)(-3)) - (y - 2)(3(2) - (-4)(2)) + (z + 3)(3(-3) - 2(2)) = 0$
$(x + 1)(4 - 12) - (y - 2)(6 + 8) + (z + 3)(-9 - 4) = 0$
$-8(x + 1) - 14(y - 2) - 13(z + 3) = 0$
$-8x - 8 - 14y + 28 - 13z - 39 = 0$
$-8x - 14y - 13z - 19 = 0$
Multiplying by $-1$,we get $8x + 14y + 13z + 19 = 0$.

(C) The equation of the plane passing through $(2,1,0)$,$(4,1,1)$,and $(5,0,1)$ is given by the determinant equation:
$\begin{vmatrix} x-2 & y-1 & z-0 \\ 4-2 & 1-1 & 1-0 \\ 5-2 & 0-1 & 1-0 \end{vmatrix} = 0$
$\Rightarrow \begin{vmatrix} x-2 & y-1 & z \\ 2 & 0 & 1 \\ 3 & -1 & 1 \end{vmatrix} = 0$
Expanding along the first row:
$(x-2)(0 - (-1)) - (y-1)(2 - 3) + z(-2 - 0) = 0$
$(x-2)(1) - (y-1)(-1) - 2z = 0$
$x - 2 + y - 1 - 2z = 0$
$x + y - 2z = 3$
Let $R'(x, y, z)$ be the image of $R(2, 1, 6)$ with respect to the plane $x + y - 2z - 3 = 0$.
The formula for the image $(x, y, z)$ of a point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d = 0$ is:
$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = -2 \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2}$
Substituting the values:
$\frac{x-2}{1} = \frac{y-1}{1} = \frac{z-6}{-2} = -2 \frac{2 + 1 - 2(6) - 3}{1^2 + 1^2 + (-2)^2}$
$\frac{x-2}{1} = \frac{y-1}{1} = \frac{z-6}{-2} = -2 \frac{3 - 12 - 3}{1 + 1 + 4} = -2 \frac{-12}{6} = 4$
Equating each part to $4$:
$x - 2 = 4 \Rightarrow x = 6$
$y - 1 = 4 \Rightarrow y = 5$
$z - 6 = -8 \Rightarrow z = -2$
Therefore,the image $R'$ is $(6, 5, -2)$.

(B) The distance of a point $(x_1, y_1, z_1)$ from 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 that the points $(1, -1, \lambda)$ and $(-3, 0, 1)$ are equidistant from the plane $3x - 4y - 12z + 13 = 0$,we have:
$\frac{|3(1) - 4(-1) - 12(\lambda) + 13|}{\sqrt{3^2 + (-4)^2 + (-12)^2}} = \frac{|3(-3) - 4(0) - 12(1) + 13|}{\sqrt{3^2 + (-4)^2 + (-12)^2}}$
$|3 + 4 - 12\lambda + 13| = |-9 - 0 - 12 + 13|$
$|20 - 12\lambda| = |-8|$
$|20 - 12\lambda| = 8$
This implies $20 - 12\lambda = 8$ or $20 - 12\lambda = -8$.
Case $1$: $20 - 12\lambda = 8 \Rightarrow 12\lambda = 12 \Rightarrow \lambda = 1$.
Case $2$: $20 - 12\lambda = -8 \Rightarrow 12\lambda = 28 \Rightarrow \lambda = \frac{28}{12} = \frac{7}{3}$.
The sum of all possible values of $\lambda$ is $1 + \frac{7}{3} = \frac{10}{3}$.

(A) The normal vector $\vec{n}$ is perpendicular to the direction vectors of the two lines,$\vec{v_1} = (1, -2, 3)$ and $\vec{v_2} = (2, -1, -1)$.
$\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{vmatrix} = \hat{i}(2+3) - \hat{j}(-1-6) + \hat{k}(-1+4) = 5\hat{i} + 7\hat{j} + 3\hat{k}$.
The equation of the plane passing through $(1, -1, -1)$ with normal vector $(5, 7, 3)$ is $5(x-1) + 7(y+1) + 3(z+1) = 0$,which simplifies to $5x + 7y + 3z + 5 = 0$.
The distance $d$ of the point $(1, 3, -7)$ from the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Substituting the values: $d = \frac{|5(1) + 7(3) + 3(-7) + 5|}{\sqrt{5^2 + 7^2 + 3^2}} = \frac{|5 + 21 - 21 + 5|}{\sqrt{25 + 49 + 9}} = \frac{10}{\sqrt{83}}$ units.

(C) The equation of a plane passing through $(1, 1, 1)$ is given by $a(x - 1) + b(y - 1) + c(z - 1) = 0$.
Since this plane is perpendicular to the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$,its normal vector $\vec{n} = (a, b, c)$ must be perpendicular to the normal vectors of the given planes,$\vec{n_1} = (2, 1, -2)$ and $\vec{n_2} = (3, -6, -2)$.
Thus,the normal vector $\vec{n}$ is given by the cross product $\vec{n_1} \times \vec{n_2}$:
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 3 & -6 & -2 \end{vmatrix} = \hat{i}(-2 - 12) - \hat{j}(-4 + 6) + \hat{k}(-12 - 3) = -14\hat{i} - 2\hat{j} - 15\hat{k}$.
Taking the normal vector as $(14, 2, 15)$,the equation of the plane is $14(x - 1) + 2(y - 1) + 15(z - 1) = 0$.
Expanding this,we get $14x - 14 + 2y - 2 + 15z - 15 = 0$,which simplifies to $14x + 2y + 15z = 31$.