MHT CET 2022 Mathematics Question Paper with Answer and Solution

(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 $Var(X) = E(X^2) - [E(X)]^2 = \frac{17}{2} - (\frac{5}{2})^2 = \frac{17}{2} - \frac{25}{4} = \frac{34-25}{4} = \frac{9}{4}$.

(C) For a function to be a probability mass function (p.m.f.),the sum of all probabilities must be equal to $1$.
$\sum_{x=0}^{3} P(X=x) = 1$
$P(0) + P(1) + P(2) + P(3) = 1$
$K \cdot \frac{2^0}{0!} + K \cdot \frac{2^1}{1!} + K \cdot \frac{2^2}{2!} + K \cdot \frac{2^3}{3!} = 1$
$K \cdot (\frac{1}{1} + \frac{2}{1} + \frac{4}{2} + \frac{8}{6}) = 1$
$K \cdot (1 + 2 + 2 + \frac{4}{3}) = 1$
$K \cdot (5 + \frac{4}{3}) = 1$
$K \cdot (\frac{15+4}{3}) = 1$
$K \cdot \frac{19}{3} = 1$
$K = \frac{3}{19}$

(C) Let $X$ be the random variable representing the number of kings drawn.
Since the cards are drawn with replacement,the trials are independent.
Total number of cards = $52$.
Number of kings = $4$.
Probability of drawing a king in a single trial,$p = \frac{4}{52} = \frac{1}{13}$.
Probability of not drawing a king in a single trial,$q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.
Since there are $n = 2$ trials,$X$ follows a binomial distribution $B(n, p) = B(2, \frac{1}{13})$.
The probability distribution is given by $P(X = k) = {}^nC_k p^k q^{n-k}$.
For $X = 0$: $P(X = 0) = {}^2C_0 (\frac{1}{13})^0 (\frac{12}{13})^2 = 1 \cdot 1 \cdot \frac{144}{169} = \frac{144}{169}$.
For $X = 1$: $P(X = 1) = {}^2C_1 (\frac{1}{13})^1 (\frac{12}{13})^1 = 2 \cdot \frac{1}{13} \cdot \frac{12}{13} = \frac{24}{169}$.
For $X = 2$: $P(X = 2) = {}^2C_2 (\frac{1}{13})^2 (\frac{12}{13})^0 = 1 \cdot \frac{1}{169} \cdot 1 = \frac{1}{169}$.
Thus,the probability distribution is:
$X=0, P(X)=\frac{144}{169}$
$X=1, P(X)=\frac{24}{169}$
$X=2, P(X)=\frac{1}{169}$
Therefore,option $(C)$ is correct.

(B) The variance of a random variable $X$ is given by $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$.
First,we calculate $E(X) = \sum_{x=1}^{10} x P(X=x) = \frac{1}{10} (1 + 2 + 3 + \ldots + 10) = \frac{1}{10} \times \frac{10 \times 11}{2} = 5.5$.
Next,we calculate $E(X^2) = \sum_{x=1}^{10} x^2 P(X=x) = \frac{1}{10} (1^2 + 2^2 + 3^2 + \ldots + 10^2) = \frac{1}{10} \times \frac{10 \times 11 \times 21}{6} = \frac{231}{6} = 38.5$.
Finally,$\operatorname{Var}(X) = 38.5 - (5.5)^2 = 38.5 - 30.25 = 8.25$.
Since $8.25 = \frac{33}{4}$,the correct option is $B$.

(A) The expected daily demand (mean) is given by $E(X) = \Sigma P_i x_i$.
$E(X) = (5 \times 0.07) + (6 \times 0.2) + (7 \times 0.3) + (8 \times 0.3) + (9 \times 0.07) + (10 \times 0.06)$
$E(X) = 0.35 + 1.2 + 2.1 + 2.4 + 0.63 + 0.6 = 7.28$.
The variance is given by $Var(X) = \Sigma P_i x_i^2 - (E(X))^2$.
First,calculate $\Sigma P_i x_i^2$:
$\Sigma P_i x_i^2 = (25 \times 0.07) + (36 \times 0.2) + (49 \times 0.3) + (64 \times 0.3) + (81 \times 0.07) + (100 \times 0.06)$
$\Sigma P_i x_i^2 = 1.75 + 7.2 + 14.7 + 19.2 + 5.67 + 6.0 = 54.52$.
Now,calculate variance:
$Var(X) = 54.52 - (7.28)^2$
$Var(X) = 54.52 - 52.9984 \approx 1.52$.
Thus,the expected demand is $7.28$ and the variance is $1.52$.

(B) Let the points be $\vec{p} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{q} = -\hat{i} + \hat{j} + \hat{k}$.
For external division in the ratio $m:n = 1:2$,the position vector $\vec{r}$ is given by the formula:
$\vec{r} = \frac{m\vec{q} - n\vec{p}}{m - n}$
Substituting the values:
$\vec{r} = \frac{1(-\hat{i} + \hat{j} + \hat{k}) - 2(\hat{i} + 2\hat{j} - \hat{k})}{1 - 2}$
$\vec{r} = \frac{-\hat{i} + \hat{j} + \hat{k} - 2\hat{i} - 4\hat{j} + 2\hat{k}}{-1}$
$\vec{r} = \frac{-3\hat{i} - 3\hat{j} + 3\hat{k}}{-1}$
$\vec{r} = 3\hat{i} + 3\hat{j} - 3\hat{k}$

(D) Let the direction cosines of the line be $l, m, n$. Since the line makes angles $\frac{\alpha}{2}, \frac{\beta}{2}, \frac{\gamma}{2}$ with the axes,we have $l = \cos(\frac{\alpha}{2}), m = \cos(\frac{\beta}{2}), n = \cos(\frac{\gamma}{2})$.
Since $l^2 + m^2 + n^2 = 1$,we have $\cos^2(\frac{\alpha}{2}) + \cos^2(\frac{\beta}{2}) + \cos^2(\frac{\gamma}{2}) = 1$.
We know that $\cos \theta = 2 \cos^2(\frac{\theta}{2}) - 1$,so $\cos^2(\frac{\theta}{2}) = \frac{1 + \cos \theta}{2}$.
Substituting this into the identity:
$\frac{1 + \cos \alpha}{2} + \frac{1 + \cos \beta}{2} + \frac{1 + \cos \gamma}{2} = 1$
$\frac{3 + \cos \alpha + \cos \beta + \cos \gamma}{2} = 1$
$3 + \cos \alpha + \cos \beta + \cos \gamma = 2$
$\cos \alpha + \cos \beta + \cos \gamma = 2 - 3 = -1$.

(B) Let the direction cosines of the line be $l, m, n$. Since the line makes the same angle $\alpha$ with the $x$ and $y$ axes,we have $l = \cos \alpha$ and $m = \cos \alpha$. Let the angle with the $z$-axis be $\theta$,so $n = \cos \theta$.
We know that $l^2 + m^2 + n^2 = 1$,which implies $\cos^2 \alpha + \cos^2 \alpha + \cos^2 \theta = 1$.
Using the identity $\cos^2 \phi = 1 - \sin^2 \phi$,we get $(1 - \sin^2 \alpha) + (1 - \sin^2 \alpha) + (1 - \sin^2 \theta) = 1$.
$3 - 2 \sin^2 \alpha - \sin^2 \theta = 1$.
Given $\sin^2 \theta = 2 \sin^2 \alpha$,substituting this into the equation:
$3 - 2 \sin^2 \alpha - 2 \sin^2 \alpha = 1$.
$3 - 4 \sin^2 \alpha = 1$.
$4 \sin^2 \alpha = 2$.
$\sin^2 \alpha = \frac{1}{2} = \sin^2 \frac{\pi}{4}$.
Thus,$\alpha = \frac{\pi}{4}$.

(A) Let $\vec{a} = 2\hat{i}+3\hat{j}+\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = \hat{i}+2\hat{j}+3\hat{k}$.
The vector perpendicular to the plane containing $\vec{b}$ and $\vec{c}$ is given by $\vec{n} = \vec{b} \times \vec{c}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i} - 2\hat{j} + \hat{k}$.
The projection of $\vec{a}$ on $\vec{n}$ is given by $\frac{|\vec{a} \cdot \vec{n}|}{|\vec{n}|}$.
$\vec{a} \cdot \vec{n} = (2\hat{i}+3\hat{j}+\hat{k}) \cdot (\hat{i}-2\hat{j}+\hat{k}) = (2)(1) + (3)(-2) + (1)(1) = 2 - 6 + 1 = -3$.
$|\vec{n}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}$.
Magnitude of projection = $\frac{|-3|}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \frac{3}{\sqrt{2} \times \sqrt{3}} = \frac{\sqrt{3}}{\sqrt{2}} = \sqrt{\frac{3}{2}}$ units.

(C) The vector projection of $\vec{b}$ on $\vec{a}$ is given by the formula: $\left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right) \vec{a}$.
First,calculate the dot product $\vec{b} \cdot \vec{a} = (7)(3) + (-5)(2) + (-1)(5) = 21 - 10 - 5 = 6$.
Next,calculate the square of the magnitude of $\vec{a}$: $|\vec{a}|^2 = 3^2 + 2^2 + 5^2 = 9 + 4 + 25 = 38$.
Now,substitute these values into the formula:
Vector projection = $\frac{6}{38} (3 \hat{i} + 2 \hat{j} + 5 \hat{k}) = \frac{3}{19} (3 \hat{i} + 2 \hat{j} + 5 \hat{k})$.

(A) 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:
$\cos \theta = \frac{|(2)(1) + (2)(2) + (-1)(2)|}{\sqrt{2^2 + 2^2 + (-1)^2} \sqrt{1^2 + 2^2 + 2^2}}$
$\cos \theta = \frac{|2 + 4 - 2|}{\sqrt{4 + 4 + 1} \sqrt{1 + 4 + 4}}$
$\cos \theta = \frac{4}{\sqrt{9} \sqrt{9}} = \frac{4}{3 \times 3} = \frac{4}{9}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{4}{9}\right)$.

(B) Let the direction ratios of the two lines be $\vec{v_1} = (2, -3, 1)$ and $\vec{v_2} = (1, 2, -2)$.
Since the required line is perpendicular to both lines,its direction ratios $(a, b, c)$ are given by the cross product $\vec{v_1} \times \vec{v_2}$.
$\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(6-2) - \hat{j}(-4-1) + \hat{k}(4+3) = 4\hat{i} + 5\hat{j} + 7\hat{k}$.
Thus,the direction ratios are $(4, 5, 7)$.
The magnitude is $\sqrt{4^2 + 5^2 + 7^2} = \sqrt{16 + 25 + 49} = \sqrt{90}$.
The direction cosines are $\pm \frac{4}{\sqrt{90}}, \pm \frac{5}{\sqrt{90}}, \pm \frac{7}{\sqrt{90}}$.

(B) Let $\vec{a} = 2\hat{i}+\hat{j}+\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = \hat{i}+2\hat{j}+3\hat{k}$.
The vector perpendicular to the plane containing $\vec{b}$ and $\vec{c}$ is given by $\vec{n} = \vec{b} \times \vec{c}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i} - 2\hat{j} + \hat{k}$.
The magnitude of the projection of $\vec{a}$ on $\vec{n}$ is given by $\left| \frac{\vec{a} \cdot \vec{n}}{|\vec{n}|} \right|$.
$\vec{a} \cdot \vec{n} = (2)(1) + (1)(-2) + (1)(1) = 2 - 2 + 1 = 1$.
$|\vec{n}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.
Therefore,the required projection is $\frac{1}{\sqrt{6}}$ units.

(B) Let the direction ratios of the required line be $a, b, c$. Since the line is perpendicular to the lines with direction ratios $\langle -1, 2, 2 \rangle$ and $\langle 0, 2, 1 \rangle$,the vector $\vec{n} = \langle a, b, c \rangle$ is parallel to the cross product of the two given vectors.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ 0 & 2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(-1-0) + \hat{k}(-2-0) = -2\hat{i} + \hat{j} - 2\hat{k}$.
Thus,the direction ratios are $\langle -2, 1, -2 \rangle$ or $\langle 2, -1, 2 \rangle$.
The magnitude of the vector is $\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.
The direction cosines are $\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \rangle$.

(B) 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:
$a + 2b + 3c = 0$ --- $(i)$
$-3a + 2b + 5c = 0$ --- $(ii)$
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 line passes through the point $(-1, 3, -2)$.
Therefore,the equation of the line is $\frac{x - (-1)}{2} = \frac{y - 3}{-7} = \frac{z - (-2)}{4}$,which simplifies to $\frac{x+1}{2} = \frac{y-3}{-7} = \frac{z+2}{4}$.

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

(C) The given lines are $x = -y, z = 0$ and $x = 0, z = 0$.
We can write these lines in symmetric form:
Line $1$: $\frac{x}{1} = \frac{y}{-1} = \frac{z}{0}$,so the direction vector is $\vec{v_1} = (1, -1, 0)$.
Line $2$: $\frac{x}{0} = \frac{y}{1} = \frac{z}{0}$,so the direction vector is $\vec{v_2} = (0, 1, 0)$.
The angle $\theta$ between the lines is given by $\cos \theta = \left| \frac{\vec{v_1} \cdot \vec{v_2}}{|\vec{v_1}| |\vec{v_2}|} \right|$.
Calculating the dot product: $\vec{v_1} \cdot \vec{v_2} = (1)(0) + (-1)(1) + (0)(0) = -1$.
Calculating the magnitudes: $|\vec{v_1}| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2}$ and $|\vec{v_2}| = \sqrt{0^2 + 1^2 + 0^2} = 1$.
Thus,$\cos \theta = \left| \frac{-1}{\sqrt{2} \times 1} \right| = \frac{1}{\sqrt{2}}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$.

(A) The line passes through the point with position vector $\vec{a} = 2 \hat{i} + \hat{j} - 3 \hat{k}$.
Since the line is perpendicular to the vectors $\vec{b}_1 = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b}_2 = \hat{i} + 2 \hat{j} - \hat{k}$,its direction vector $\vec{v}$ is given by the cross product $\vec{b}_1 \times \vec{b}_2$.
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(-1 - 1) + \hat{k}(2 - 1) = -3 \hat{i} + 2 \hat{j} + \hat{k}$.
The vector equation of the line is $\vec{r} = \vec{a} + \lambda \vec{v}$.
Substituting the values,we get $\vec{r} = (2 \hat{i} + \hat{j} - 3 \hat{k}) + \lambda(-3 \hat{i} + 2 \hat{j} + \hat{k})$.

(A) Let the $yz$-plane divide the line segment joining the points $A(3, 5, -7)$ and $B(-2, 1, 8)$ in the ratio $\lambda : 1$.
Since the point lies on the $yz$-plane,its $x$-coordinate must be $0$.
Using the section formula for the $x$-coordinate:
$\frac{\lambda(-2) + 1(3)}{\lambda + 1} = 0$
$-2\lambda + 3 = 0 \Rightarrow \lambda = \frac{3}{2}$.
Thus,the ratio is $3 : 2$.
Now,find the $y$ and $z$ coordinates using the ratio $3 : 2$:
$y = \frac{3(1) + 2(5)}{3 + 2} = \frac{3 + 10}{5} = \frac{13}{5}$
$z = \frac{3(8) + 2(-7)}{3 + 2} = \frac{24 - 14}{5} = \frac{10}{5} = 2$
Therefore,the required point is $\left(0, \frac{13}{5}, 2\right)$.

(C) Let the given vectors be $\vec{a} = \hat{i}-2\hat{j}+\hat{k}$ and $\vec{b} = 2\hat{i}+\hat{j}-\hat{k}$.
Since the line is perpendicular to both vectors,its direction vector $\vec{v}$ is given by the cross product $\vec{a} \times \vec{b}$.
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 2 & 1 & -1 \end{vmatrix} = \hat{i}(2-1) - \hat{j}(-1-2) + \hat{k}(1+4) = 1\hat{i} + 3\hat{j} + 5\hat{k}$.
The line passes through $(-3, 0, 1)$ with direction ratios $(1, 3, 5)$.
The Cartesian equation is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
Substituting the values,we get $\frac{x - (-3)}{1} = \frac{y - 0}{3} = \frac{z - 1}{5}$,which simplifies to $\frac{x+3}{1} = \frac{y}{3} = \frac{z-1}{5}$.

(C) The magnitude of a vector remains invariant under the rotation of the coordinate system about the origin.
Therefore,the magnitude of $\vec{a}$ before rotation is equal to the magnitude of $\vec{a}$ after rotation.
$|\vec{a}|^2 = (2p)^2 + 1^2 = (p+1)^2 + 1^2$
$4p^2 + 1 = p^2 + 2p + 1 + 1$
$4p^2 + 1 = p^2 + 2p + 2$
$3p^2 - 2p - 1 = 0$
Factoring the quadratic equation:
$3p^2 - 3p + p - 1 = 0$
$3p(p-1) + 1(p-1) = 0$
$(3p+1)(p-1) = 0$
Thus,$p=1$ or $p=-\frac{1}{3}$.

(B) The given lines are $L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$ and $L_2: \vec{r} = (0\hat{i} + 0\hat{j} + 0\hat{k}) + \mu(2\hat{i} - 2\hat{j} + \hat{k})$.
Here,$\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{a_2} = 0$,and $\vec{b} = 2\hat{i} - 2\hat{j} + \hat{k}$.
The distance $d$ between two parallel lines is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|}$.
First,$\vec{a_2} - \vec{a_1} = -\hat{i} - 2\hat{j} - 3\hat{k}$.
Now,$(\vec{a_2} - \vec{a_1}) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -2 & -3 \\ 2 & -2 & 1 \end{vmatrix} = \hat{i}(-2 - 6) - \hat{j}(-1 + 6) + \hat{k}(2 + 4) = -8\hat{i} - 5\hat{j} + 6\hat{k}$.
The magnitude is $|(\vec{a_2} - \vec{a_1}) \times \vec{b}| = \sqrt{(-8)^2 + (-5)^2 + 6^2} = \sqrt{64 + 25 + 36} = \sqrt{125} = 5\sqrt{5}$.
The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Thus,the distance $d = \frac{5\sqrt{5}}{3}$ units.

(C) The normal vector $\vec{n}$ to the plane is along $\overrightarrow{AB}$.
$\overrightarrow{AB} = \vec{B} - \vec{A} = (\hat{i}-2 \hat{j}-4 \hat{k}) - (3 \hat{i}+\hat{j}+2 \hat{k}) = -2 \hat{i}-3 \hat{j}-6 \hat{k}$.
Since the plane is perpendicular to $\overrightarrow{AB}$,the normal vector to the plane can be taken as $\vec{n} = 2 \hat{i}+3 \hat{j}+6 \hat{k}$.
The equation of a plane passing through a point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = a \hat{i}+b \hat{j}+c \hat{k}$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Substituting the point $B(1, -2, -4)$ and the normal vector components $(2, 3, 6)$:
$2(x-1) + 3(y+2) + 6(z+4) = 0$.
$2x - 2 + 3y + 6 + 6z + 24 = 0$.
$2x + 3y + 6z + 28 = 0$.

(B) Let the $yz$-plane divide the line segment joining the points $(a, 1, 6)$ and $(3, 4, b)$ in the ratio $\lambda : 1$.
Using the section formula,the point of intersection is given by:
$\left(\frac{3\lambda + a}{\lambda + 1}, \frac{4\lambda + 1}{\lambda + 1}, \frac{b\lambda + 6}{\lambda + 1}\right) = \left(0, \frac{17}{2}, \frac{-13}{2}\right)$.
Equating the coordinates:
$1) \frac{3\lambda + a}{\lambda + 1} = 0 \Rightarrow 3\lambda + a = 0 \Rightarrow a = -3\lambda$.
$2) \frac{4\lambda + 1}{\lambda + 1} = \frac{17}{2} \Rightarrow 8\lambda + 2 = 17\lambda + 17 \Rightarrow -9\lambda = 15 \Rightarrow \lambda = -\frac{15}{9} = -\frac{5}{3}$.
Substituting $\lambda = -\frac{5}{3}$ into $a = -3\lambda$:
$a = -3\left(-\frac{5}{3}\right) = 5$.
$3) \frac{b\lambda + 6}{\lambda + 1} = -\frac{13}{2} \Rightarrow \frac{b(-\frac{5}{3}) + 6}{-\frac{5}{3} + 1} = -\frac{13}{2} \Rightarrow \frac{-\frac{5b}{3} + 6}{-\frac{2}{3}} = -\frac{13}{2}$.
Multiply both sides by $-\frac{2}{3}$:
$-\frac{5b}{3} + 6 = \left(-\frac{13}{2}\right) \times \left(-\frac{2}{3}\right) = \frac{13}{3}$.
$-\frac{5b}{3} = \frac{13}{3} - 6 = \frac{13 - 18}{3} = -\frac{5}{3}$.
$b = 1$.
Thus,$a = 5$ and $b = 1$.

(C) The normal vector $\vec{n}$ of the required plane is perpendicular to the normals of the given planes $\vec{n}_1 = 2\hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{n}_2 = \hat{i} + 2\hat{j} - 3\hat{k}$.
Thus,$\vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -2 \\ 1 & 2 & -3 \end{vmatrix} = \hat{i}(-9 + 4) - \hat{j}(-6 + 2) + \hat{k}(4 - 3) = -5\hat{i} + 4\hat{j} + \hat{k}$.
The equation of the plane passing through $(x_0, y_0, z_0) = (1, -1, 2)$ with normal $\vec{n} = -5\hat{i} + 4\hat{j} + \hat{k}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
$\vec{a} \cdot \vec{n} = (1\hat{i} - 1\hat{j} + 2\hat{k}) \cdot (-5\hat{i} + 4\hat{j} + \hat{k}) = (1)(-5) + (-1)(4) + (2)(1) = -5 - 4 + 2 = -7$.
Therefore,the vector equation is $\vec{r} \cdot (-5\hat{i} + 4\hat{j} + \hat{k}) = -7$.

(A) The direction vectors of the two lines are $\vec{b_1} = 2\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b_2} = \hat{i} + 2\hat{j} + 2\hat{k}$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (2)(1) + (-2)(2) + (1)(2) = 2 - 4 + 2 = 0$.
Since the dot product is $0$,the angle $\theta$ is $\cos^{-1}(0) = 90^{\circ}$.

(A) The Cartesian equation of a line is given by $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$.
Comparing this with the given equation $\frac{x-(-2)}{3}=\frac{y-4}{2}=\frac{z-5}{5}$,we identify the point $(x_1, y_1, z_1) = (-2, 4, 5)$ through which the line passes.
The direction ratios $(a, b, c)$ are $(3, 2, 5)$.
The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$.
Here,$\vec{a} = -2 \hat{i} + 4 \hat{j} + 5 \hat{k}$ and $\vec{b} = 3 \hat{i} + 2 \hat{j} + 5 \hat{k}$.
Thus,the vector equation is $\vec{r} = (-2 \hat{i} + 4 \hat{j} + 5 \hat{k}) + \lambda(3 \hat{i} + 2 \hat{j} + 5 \hat{k})$.

(NONE) The given lines are parallel because their direction ratios are the same: $(2, -1, 2)$.
Let the lines be $L_1: \vec{r} = 0\hat{i} + 0\hat{j} + 0\hat{k} + \lambda(2\hat{i} - \hat{j} + 2\hat{k})$ and $L_2: \vec{r} = 1\hat{i} + 1\hat{j} + 2\hat{k} + \mu(2\hat{i} - \hat{j} + 2\hat{k})$.
Here,$\vec{a_1} = (0, 0, 0)$,$\vec{a_2} = (1, 1, 2)$,and $\vec{b} = (2, -1, 2)$.
The distance $d$ between two parallel lines is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|}$.
$\vec{a_2} - \vec{a_1} = (1-0)\hat{i} + (1-0)\hat{j} + (2-0)\hat{k} = \hat{i} + \hat{j} + 2\hat{k}$.
$(\vec{a_2} - \vec{a_1}) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & -1 & 2 \end{vmatrix} = \hat{i}(2 - (-2)) - \hat{j}(2 - 4) + \hat{k}(-1 - 2) = 4\hat{i} + 2\hat{j} - 3\hat{k}$.
Magnitude $|(\vec{a_2} - \vec{a_1}) \times \vec{b}| = \sqrt{4^2 + 2^2 + (-3)^2} = \sqrt{16 + 4 + 9} = \sqrt{29}$.
Magnitude $|\vec{b}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
Thus,$d = \frac{\sqrt{29}}{3}$ units. Note: The provided options do not match the calculated result. Based on standard evaluation,the correct distance is $\frac{\sqrt{29}}{3}$.

(B) Let the given line be $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda$.
Any point on this line can be represented as $M(5\lambda-3, 2\lambda+1, 3\lambda-4)$.
Let $P$ be the point $(0,2,3)$. The vector $\vec{PM}$ is given by $(5\lambda-3-0, 2\lambda+1-2, 3\lambda-4-3) = (5\lambda-3, 2\lambda-1, 3\lambda-7)$.
Since $PM$ is perpendicular to the line,the dot product of $\vec{PM}$ and the direction vector of the line $\vec{v} = (5, 2, 3)$ must be zero.
$5(5\lambda-3) + 2(2\lambda-1) + 3(3\lambda-7) = 0$
$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0$
$38\lambda - 38 = 0$
$\lambda = 1$.
Substituting $\lambda = 1$ in the coordinates of $M$,we get:
$x = 5(1)-3 = 2$
$y = 2(1)+1 = 3$
$z = 3(1)-4 = -1$
Thus,the foot of the perpendicular is $(2,3,-1)$.

(B) To check if the lines intersect,we calculate the shortest distance ($S$.$D$.) between them. The lines are given by $\vec{r_1} = (1, -1, 1) + \lambda(3, 2, 5)$ and $\vec{r_2} = (-2, 1, -1) + \mu(4, 3, -2)$.
The shortest distance is given by the formula $S.D. = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$.
Here,$\vec{a_2} - \vec{a_1} = (-2-1, 1-(-1), -1-1) = (-3, 2, -2)$.
The cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 5 \\ 4 & 3 & -2 \end{vmatrix} = \hat{i}(-4-15) - \hat{j}(-6-20) + \hat{k}(9-8) = -19\hat{i} + 26\hat{j} + 1\hat{k}$.
Now,$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-3)(-19) + (2)(26) + (-2)(1) = 57 + 52 - 2 = 107$.
Since the shortest distance is $\frac{107}{\sqrt{(-19)^2 + 26^2 + 1^2}} = \frac{107}{\sqrt{361 + 676 + 1}} = \frac{107}{\sqrt{1038}} \neq 0$,the lines do not intersect.

(B) Let the given line be $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = \lambda$.
Any point $P$ on this line is given by $(5\lambda - 3, 2\lambda + 1, 3\lambda - 4)$.
Let $A = (0, 2, 3)$ be the given point. The direction ratios of the line $AP$ are $(5\lambda - 3 - 0, 2\lambda + 1 - 2, 3\lambda - 4 - 3)$,which simplifies to $(5\lambda - 3, 2\lambda - 1, 3\lambda - 7)$.
Since $AP$ is perpendicular to the given line with direction ratios $(5, 2, 3)$,their dot product must be zero:
$5(5\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 7) = 0$.
$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0$.
$38\lambda - 38 = 0 \Rightarrow \lambda = 1$.
The foot of the perpendicular is $P = (5(1) - 3, 2(1) + 1, 3(1) - 4) = (2, 3, -1)$.
The length of the perpendicular $AP$ is $\sqrt{(2-0)^2 + (3-2)^2 + (-1-3)^2} = \sqrt{2^2 + 1^2 + (-4)^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$ units.

(B) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.
Substituting the given points $(3, 1, 2)$ and $(-1, 2, 1)$ into the formula:
$\frac{x-3}{-1-3} = \frac{y-1}{2-1} = \frac{z-2}{1-2}$
$\frac{x-3}{-4} = \frac{y-1}{1} = \frac{z-2}{-1}$

(A) The normal vector $\vec{n}$ to the plane is the vector from the origin $O(0, 0, 0)$ to the foot of the perpendicular $M(-1, -2, 2)$.
Thus,$\vec{n} = \vec{OM} = -\hat{i} - 2\hat{j} + 2\hat{k}$.
The equation of a plane passing through a point $M$ with normal vector $\vec{n}$ is given by $\vec{r} \cdot \vec{n} = \vec{OM} \cdot \vec{n}$.
Here,$\vec{OM} \cdot \vec{n} = (-1, -2, 2) \cdot (-1, -2, 2) = (-1)^2 + (-2)^2 + (2)^2 = 1 + 4 + 4 = 9$.
Therefore,the vector equation of the plane is $\vec{r} \cdot(-\hat{i} - 2\hat{j} + 2\hat{k}) = 9$.

(D) The lines are given by $\vec{r_1} = (3, 8, 3) + \lambda(3, -1, 1)$ and $\vec{r_2} = (-3, -7, 6) + \mu(-3, 2, 4)$.
The shortest distance $d$ between two lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{ |\vec{b_1} \times \vec{b_2}| }$.
Here,$\vec{a_2} - \vec{a_1} = (-3-3, -7-8, 6-3) = (-6, -15, 3)$.
The cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix} = \hat{i}(-4-2) - \hat{j}(12+3) + \hat{k}(6-3) = -6\hat{i} - 15\hat{j} + 3\hat{k}$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.
The dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-6)(-6) + (-15)(-15) + (3)(3) = 36 + 225 + 9 = 270$.
Therefore,$d = \frac{|270|}{3\sqrt{30}} = \frac{270}{3\sqrt{30}} = \frac{90}{\sqrt{30}} = 3\sqrt{30}$ 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 the plane is parallel to the lines with direction ratios $(1, 0, -1)$ and $(-1, 1, 0)$,the normal vector $(a, b, c)$ must be perpendicular to both direction vectors.
Thus,$a(1) + b(0) + c(-1) = 0 \Rightarrow a - c = 0$ and $a(-1) + b(1) + c(0) = 0 \Rightarrow -a + b = 0$.
This implies $a = c$ and $a = b$,so $a = b = c$.
Substituting $a=b=c$ into the plane equation,we get $a(x-1) + a(y-1) + a(z-1) = 0$,which simplifies to $x + y + z = 3$.
Dividing by $3$,we get the intercept form $\frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1$.
Thus,the coordinates of $A, B, C$ are $(3, 0, 0), (0, 3, 0),$ and $(0, 0, 3)$ respectively.
The volume of the tetrahedron $OABC$ is given by $V = \frac{1}{6} |x_A y_B z_C| = \frac{1}{6} |3 \times 3 \times 3| = \frac{27}{6} = \frac{9}{2}$ cubic units.

(C) The equation of any plane passing through the intersection of the planes $P_1: x+y+z-1=0$ and $P_2: 2x+3y-z+4=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+y+z-1) + \lambda(2x+3y-z+4) = 0$
$(1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (4\lambda-1) = 0$
Since the plane is parallel to the $X$-axis,its normal vector must be perpendicular to the $X$-axis (direction vector $\vec{i} = (1, 0, 0)$).
Therefore,the coefficient of $x$ must be zero: $1+2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Substituting $\lambda = -\frac{1}{2}$ into the equation:
$(1+2(-\frac{1}{2}))x + (1+3(-\frac{1}{2}))y + (1-(-\frac{1}{2}))z + (4(-\frac{1}{2})-1) = 0$
$0x + (1-\frac{3}{2})y + (1+\frac{1}{2})z + (-2-1) = 0$
$-\frac{1}{2}y + \frac{3}{2}z - 3 = 0$
Multiplying by $-2$,we get $y-3z+6=0$.

(B) The coordinates of the foot of the perpendicular $(x, y, z)$ from the origin $(0, 0, 0)$ to the plane $ax + by + cz = d$ are given by the formula:
$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \frac{-(ax_1 + by_1 + cz_1 - d)}{a^2 + b^2 + c^2}$
Here,$(x_1, y_1, z_1) = (0, 0, 0)$,$a = 2$,$b = 6$,$c = -3$,and $d = 63$.
Substituting these values:
$\frac{x-0}{2} = \frac{y-0}{6} = \frac{z-0}{-3} = \frac{-(2(0) + 6(0) - 3(0) - 63)}{2^2 + 6^2 + (-3)^2}$
$\frac{x}{2} = \frac{y}{6} = \frac{z}{-3} = \frac{-(-63)}{4 + 36 + 9}$
$\frac{x}{2} = \frac{y}{6} = \frac{z}{-3} = \frac{63}{49} = \frac{9}{7}$
Now,solving for $x, y, z$:
$x = 2 \times \frac{9}{7} = \frac{18}{7}$
$y = 6 \times \frac{9}{7} = \frac{54}{7}$
$z = -3 \times \frac{9}{7} = \frac{-27}{7}$
Thus,the coordinates are $(\frac{18}{7}, \frac{54}{7}, \frac{-27}{7})$.

(D) The equation of any plane passing through the intersection of the planes $P_1: x+2y-3z+1=0$ and $P_2: 3x-2y+4z+3=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+2y-3z+1) + \lambda(3x-2y+4z+3) = 0 \quad \dots(1)$
Since the plane passes through the point $(2,2,1)$,we substitute $x=2, y=2, z=1$ into equation $(1)$:
$(2 + 2(2) - 3(1) + 1) + \lambda(3(2) - 2(2) + 4(1) + 3) = 0$
$(2 + 4 - 3 + 1) + \lambda(6 - 4 + 4 + 3) = 0$
$4 + 9\lambda = 0$
$\lambda = -\frac{4}{9}$
Substituting $\lambda = -\frac{4}{9}$ back into equation $(1)$:
$(x+2y-3z+1) - \frac{4}{9}(3x-2y+4z+3) = 0$
$9(x+2y-3z+1) - 4(3x-2y+4z+3) = 0$
$9x + 18y - 27z + 9 - 12x + 8y - 16z - 12 = 0$
$-3x + 26y - 43z - 3 = 0$
Multiplying by $-1$,we get $3x - 26y + 43z + 3 = 0$.

(D) The equation of the line passing through the origin $(0, 0, 0)$ and perpendicular to the plane $3x + 2y + 6z - 56 = 0$ is given by $\frac{x-0}{3} = \frac{y-0}{2} = \frac{z-0}{6} = k$.
Any point on this line is $(3k, 2k, 6k)$.
Since this point lies on the plane,we substitute it into the plane equation:
$3(3k) + 2(2k) + 6(6k) = 56$
$9k + 4k + 36k = 56$
$49k = 56$
$k = \frac{56}{49} = \frac{8}{7}$
Substituting $k$ back into the coordinates $(3k, 2k, 6k)$:
$x = 3 \times \frac{8}{7} = \frac{24}{7}$
$y = 2 \times \frac{8}{7} = \frac{16}{7}$
$z = 6 \times \frac{8}{7} = \frac{48}{7}$
Thus,the foot of the perpendicular is $\left(\frac{24}{7}, \frac{16}{7}, \frac{48}{7}\right)$.

(C) The equation of the plane $OPQ$ passing through $O(0,0,0), P(1,2,1)$,and $Q(2,1,3)$ is given by the determinant:
$\begin{vmatrix} x & y & z \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{vmatrix} = 0$
$x(6-1) - y(3-2) + z(1-4) = 0 \Rightarrow 5x - y - 3z = 0$.
The normal vector to plane $OPQ$ is $\vec{n_1} = 5\hat{i} - \hat{j} - 3\hat{k}$.
The equation of the plane $PQR$ passing through $P(1,2,1), Q(2,1,3)$,and $R(-1,1,2)$ is given by:
$\begin{vmatrix} x-1 & y-2 & z-1 \\ 2-1 & 1-2 & 3-1 \\ -1-1 & 1-2 & 2-1 \end{vmatrix} = 0 \Rightarrow \begin{vmatrix} x-1 & y-2 & z-1 \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{vmatrix} = 0$
$(x-1)(-1+2) - (y-2)(1+4) + (z-1)(-1-2) = 0$
$(x-1) - 5(y-2) - 3(z-1) = 0 \Rightarrow x - 5y - 3z + 12 = 0$.
The normal vector to plane $PQR$ is $\vec{n_2} = \hat{i} - 5\hat{j} - 3\hat{k}$.
The angle $\theta$ between the two planes is given by:
$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|} = \frac{|(5)(1) + (-1)(-5) + (-3)(-3)|}{\sqrt{5^2 + (-1)^2 + (-3)^2} \sqrt{1^2 + (-5)^2 + (-3)^2}}$
$\cos \theta = \frac{|5 + 5 + 9|}{\sqrt{25+1+9} \sqrt{1+25+9}} = \frac{19}{\sqrt{35} \sqrt{35}} = \frac{19}{35}$
Therefore,$\theta = \cos^{-1}\left(\frac{19}{35}\right)$.

(D) plane parallel to the $X$-axis has the general equation $by + cz + d = 0$.
Since the plane passes through the points $(2, 3, 1)$ and $(4, -5, 3)$,these points must satisfy the equation.
For point $(2, 3, 1)$: $b(3) + c(1) + d = 0 \implies 3b + c + d = 0$.
For point $(4, -5, 3)$: $b(-5) + c(3) + d = 0 \implies -5b + 3c + d = 0$.
Subtracting the two equations: $(3b + c + d) - (-5b + 3c + d) = 0 \implies 8b - 2c = 0 \implies c = 4b$.
Substituting $c = 4b$ into the first equation: $3b + 4b + d = 0 \implies d = -7b$.
The equation becomes $by + (4b)z - 7b = 0$.
Dividing by $b$ (assuming $b \neq 0$),we get $y + 4z = 7$.
Thus,the correct option is $(D)$.

(B) Let the direction ratios of the line be $(a, b, c)$. Since the line is parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$,its direction vector is perpendicular to the normals of both planes.
The normals are $\vec{n_1} = (1, -1, 2)$ and $\vec{n_2} = (3, 1, 1)$.
The direction vector $\vec{v}$ is given by the cross product $\vec{n_1} \times \vec{n_2}$:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 2) - \hat{j}(1 - 6) + \hat{k}(1 - (-3)) = -3\hat{i} + 5\hat{j} + 4\hat{k}$.
Thus,the direction ratios are $(-3, 5, 4)$.
The line passes through $(1, 2, 3)$,so the Cartesian equation is $\frac{x-1}{-3} = \frac{y-2}{5} = \frac{z-3}{4}$.

(C) The equation of the plane is given in the form $\bar{r} \cdot \bar{n} = d$,where $\bar{n} = 3 \hat{i} - 4 \hat{j} + 12 \hat{k}$ and $d = 8$.
The length of the perpendicular from the origin to the plane $\bar{r} \cdot \bar{n} = d$ is given by the formula $p = \frac{|d|}{|\bar{n}|}$.
Here,$|\bar{n}| = \sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.
Therefore,the length of the perpendicular is $p = \frac{8}{13}$ units.

(A) 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:
$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$
Substituting the given points $(1, 2, 3)$,$(-1, 4, 2)$,and $(3, 1, 1)$:
$\begin{vmatrix} x-1 & y-2 & z-3 \\ -1-1 & 4-2 & 2-3 \\ 3-1 & 1-2 & 1-3 \end{vmatrix} = 0$
$\begin{vmatrix} x-1 & y-2 & z-3 \\ -2 & 2 & -1 \\ 2 & -1 & -2 \end{vmatrix} = 0$
Expanding along the first row:
$(x-1)[(2)(-2) - (-1)(-1)] - (y-2)[(-2)(-2) - (2)(-1)] + (z-3)[(-2)(-1) - (2)(2)] = 0$
$(x-1)[-4 - 1] - (y-2)[4 + 2] + (z-3)[2 - 4] = 0$
$-5(x-1) - 6(y-2) - 2(z-3) = 0$
$-5x + 5 - 6y + 12 - 2z + 6 = 0$
$-5x - 6y - 2z + 23 = 0$
Multiplying by $-1$,we get:
$5x + 6y + 2z - 23 = 0$

(A) Two planes $\bar{r} \cdot \bar{n}_1 = d_1$ and $\bar{r} \cdot \bar{n}_2 = d_2$ are parallel if their normal vectors $\bar{n}_1$ and $\bar{n}_2$ are proportional,i.e.,$\bar{n}_1 = k \bar{n}_2$.
Given normal vectors are $\bar{n}_1 = 2 \hat{i} - \lambda \hat{j} + \hat{k}$ and $\bar{n}_2 = 4 \hat{i} - \hat{j} + \mu \hat{k}$.
For the planes to be parallel,the ratios of the components of the normal vectors must be equal:
$\frac{2}{4} = \frac{-\lambda}{-1} = \frac{1}{\mu}$
From $\frac{2}{4} = \frac{-\lambda}{-1}$,we get $\frac{1}{2} = \lambda$,so $\lambda = \frac{1}{2}$.
From $\frac{2}{4} = \frac{1}{\mu}$,we get $\frac{1}{2} = \frac{1}{\mu}$,so $\mu = 2$.
Thus,the values are $\lambda = \frac{1}{2}$ and $\mu = 2$.

(A) The normal vector $\vec{n}$ to the plane is the vector from the origin $(0, 0, 0)$ to the foot of the perpendicular $(4, -2, -5)$.
Thus,$\vec{n} = \langle 4 - 0, -2 - 0, -5 - 0 \rangle = \langle 4, -2, -5 \rangle$.
The equation of a plane with normal $\vec{n} = \langle a, b, c \rangle$ passing through a point $(x_1, y_1, z_1)$ 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,$4x - 16 - 2y - 4 - 5z - 25 = 0$.
$4x - 2y - 5z - 45 = 0$,which simplifies to $4x - 2y - 5z = 45$.

(C) The equation of any plane passing through the intersection of the planes $P_1: x+y+z-1=0$ and $P_2: 2x+3y-z+4=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+y+z-1) + \lambda(2x+3y-z+4) = 0$
$(1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (4\lambda-1) = 0$.
Since the plane is parallel to the $x$-axis,its normal vector $\vec{n} = (1+2\lambda, 1+3\lambda, 1-\lambda)$ must be perpendicular to the $x$-axis direction vector $\vec{i} = (1, 0, 0)$.
Therefore,the dot product is zero: $(1+2\lambda)(1) + (1+3\lambda)(0) + (1-\lambda)(0) = 0$.
$1+2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Substituting $\lambda = -\frac{1}{2}$ into the equation:
$(1+2(-\frac{1}{2}))x + (1+3(-\frac{1}{2}))y + (1-(-\frac{1}{2}))z + (4(-\frac{1}{2})-1) = 0$.
$0x + (1-\frac{3}{2})y + (1+\frac{1}{2})z + (-2-1) = 0$.
$-\frac{1}{2}y + \frac{3}{2}z - 3 = 0$.
Multiplying by $-2$,we get $y - 3z + 6 = 0$.

(D) Let the vector $\bar{n}$ be $a\hat{i} + a\hat{j} + a\hat{k}$ since it makes equal angles with the coordinate axes.
Given magnitude $|\bar{n}| = 3\sqrt{3}$,so $\sqrt{a^2 + a^2 + a^2} = 3\sqrt{3} \Rightarrow \sqrt{3a^2} = 3\sqrt{3} \Rightarrow a\sqrt{3} = 3\sqrt{3} \Rightarrow a = 3$.
Thus,$\bar{n} = 3\hat{i} + 3\hat{j} + 3\hat{k}$.
The equation of a plane passing through point $\vec{a} = (1, -1, 2)$ and normal to $\bar{n}$ is given by $\vec{r} \cdot \bar{n} = \vec{a} \cdot \bar{n}$.
$\vec{r} \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) = (1\hat{i} - 1\hat{j} + 2\hat{k}) \cdot (3\hat{i} + 3\hat{j} + 3\hat{k})$.
$\vec{r} \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) = (1)(3) + (-1)(3) + (2)(3) = 3 - 3 + 6 = 6$.
Therefore,the equation is $\bar{r} \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) = 6$.

(B) The line passes through the points $A(4, -1, 2)$ and $B(-3, 2, 3)$. Since the line is perpendicular to the plane,the direction ratios of the line are the direction ratios of the normal to the plane.
Direction ratios of the line are $\langle 4 - (-3), -1 - 2, 2 - 3 \rangle = \langle 7, -3, -1 \rangle$.
The plane passes through the point $(-10, 5, 4)$.
The equation of a plane passing through $(x_0, y_0, z_0)$ with normal vector $\langle a, b, c \rangle$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$.
Substituting the values,we get:
$7(x - (-10)) - 3(y - 5) - 1(z - 4) = 0$
$7(x + 10) - 3(y - 5) - 1(z - 4) = 0$
$7x + 70 - 3y + 15 - z + 4 = 0$
$7x - 3y - z + 89 = 0$.

(B) Let the required ratio be $\lambda:1$. The position vector of the point dividing the line segment joining $\vec{a} = -2\hat{i} + 4\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} + 8\hat{k}$ in the ratio $\lambda:1$ is given by $\vec{r} = \frac{\lambda\vec{b} + 1\vec{a}}{\lambda+1}$.
Substituting the coordinates,we get $\vec{r} = \frac{\lambda(3\hat{i}-5\hat{j}+8\hat{k}) + (-2\hat{i}+4\hat{j}+7\hat{k})}{\lambda+1} = \frac{(3\lambda-2)\hat{i} + (-5\lambda+4)\hat{j} + (8\lambda+7)\hat{k}}{\lambda+1}$.
Since this point lies on the plane $\vec{r} \cdot (\hat{i}-2\hat{j}+3\hat{k}) = 17$,we substitute the components:
$\frac{(3\lambda-2)(1) + (-5\lambda+4)(-2) + (8\lambda+7)(3)}{\lambda+1} = 17$.
$(3\lambda-2) + (10\lambda-8) + (24\lambda+21) = 17(\lambda+1)$.
$37\lambda + 11 = 17\lambda + 17$.
$20\lambda = 6$.
$\lambda = \frac{6}{20} = \frac{3}{10}$.
Thus,the required ratio is $3:10$.