MHT CET 2020 Mathematics Question Paper with Answer and Solution

(C) The given lines are $\ell_{1}: \bar{r} = (\hat{i}-2\hat{j}+3\hat{k}) + t(-\hat{i}+\hat{j}-2\hat{k})$ and $\ell_{2}: \bar{r} = (\hat{i}-\hat{j}+\hat{k}) + p(\hat{i}+2\hat{j}+2\hat{k})$.
Here,$\bar{a}_{1} = \hat{i}-2\hat{j}+3\hat{k}$,$\bar{b}_{1} = -\hat{i}+\hat{j}-2\hat{k}$,$\bar{a}_{2} = \hat{i}-\hat{j}+\hat{k}$,and $\bar{b}_{2} = \hat{i}+2\hat{j}+2\hat{k}$.
First,calculate $\bar{a}_{2}-\bar{a}_{1} = (\hat{i}-\hat{j}+\hat{k}) - (\hat{i}-2\hat{j}+3\hat{k}) = \hat{j}-2\hat{k}$.
Next,calculate the cross product $\bar{b}_{1} \times \bar{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & 2 \end{vmatrix} = \hat{i}(2+4) - \hat{j}(-2+2) + \hat{k}(-2-1) = 6\hat{i} - 3\hat{k}$.
The magnitude is $|\bar{b}_{1} \times \bar{b}_{2}| = \sqrt{6^2 + (-3)^2} = \sqrt{36+9} = \sqrt{45} = 3\sqrt{5}$.
The shortest distance $d = \left| \frac{(\bar{b}_{1} \times \bar{b}_{2}) \cdot (\bar{a}_{2}-\bar{a}_{1})}{|\bar{b}_{1} \times \bar{b}_{2}|} \right| = \left| \frac{(6\hat{i}-3\hat{k}) \cdot (\hat{j}-2\hat{k})}{3\sqrt{5}} \right| = \left| \frac{0 + 0 + 6}{3\sqrt{5}} \right| = \frac{6}{3\sqrt{5}} = \frac{2}{\sqrt{5}}$ units.

(B) The given equation of the line is $3x + 4 = 4y - 1 = 1 - 4z$.
We rewrite this in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ by dividing by the coefficients of $x, y, z$:
$\frac{x + 4/3}{1/3} = \frac{y - 1/4}{1/4} = \frac{z - 1/4}{-1/4}$.
Multiplying the denominators by $12$,we get the direction ratios $(a, b, c)$ as $(4, 3, -3)$.
The equation of a line passing through $(x_0, y_0, z_0) = (2, 4, 6)$ with direction ratios $(4, 3, -3)$ is:
$\frac{x-2}{4} = \frac{y-4}{3} = \frac{z-6}{-3}$.

(C) First,rewrite the equations of the lines in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
For the first line: $\frac{x-1}{5} = \frac{y+1}{3} = \frac{z-3}{-\lambda}$. The direction ratios are $\vec{v_1} = (5, 3, -\lambda)$.
For the second line: $\frac{x+1}{4} = \frac{y-1/3}{-5} = \frac{z+1}{1}$. The direction ratios are $\vec{v_2} = (4, -5, 1)$.
Since the lines are perpendicular,the dot product of their direction ratios must be zero: $\vec{v_1} \cdot \vec{v_2} = 0$.
$5(4) + 3(-5) + (-\lambda)(1) = 0$.
$20 - 15 - \lambda = 0$.
$5 - \lambda = 0$.
Therefore,$\lambda = 5$.

(A) The given lines are $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4} = \lambda$ and $L_2: \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1} = \mu$.
Any point on $L_1$ is $P(2\lambda+1, 3\lambda-1, 4\lambda+1)$ and any point on $L_2$ is $Q(\mu+3, 2\mu+k, \mu)$.
For the lines to intersect,there must exist $\lambda$ and $\mu$ such that $P=Q$.
Equating coordinates:
$2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2$ $(i)$
$3\lambda-1 = 2\mu+k \implies 3\lambda - 2\mu = k+1$ (ii)
$4\lambda+1 = \mu \implies 4\lambda - \mu = -1$ (iii)
Subtracting $(i)$ from (iii): $(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2 \implies 2\lambda = -3 \implies \lambda = -\frac{3}{2}$.
Substitute $\lambda = -\frac{3}{2}$ into $(i)$: $2(-\frac{3}{2}) - \mu = 2 \implies -3 - \mu = 2 \implies \mu = -5$.
Now substitute $\lambda = -\frac{3}{2}$ and $\mu = -5$ into (ii): $3(-\frac{3}{2}) - 2(-5) = k+1 \implies -\frac{9}{2} + 10 = k+1 \implies k = 9 - \frac{9}{2} = \frac{9}{2}$.

(D) The direction vectors of the two lines are $\vec{b_1} = \hat{i} + 2 \hat{j} + m \hat{k}$ and $\vec{b_2} = 2 \hat{i} + \hat{j} + 6 \hat{k}$.
Since the lines are perpendicular,the dot product of their direction vectors must be zero,i.e.,$\vec{b_1} \cdot \vec{b_2} = 0$.
$(1)(2) + (2)(1) + (m)(6) = 0$.
$2 + 2 + 6m = 0$.
$4 + 6m = 0$.
$6m = -4$.
$m = \frac{-4}{6} = \frac{-2}{3}$.

(A) 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 $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.
Given points are $(3, 4, -7)$ and $(6, -1, 1)$.
Substituting these values into the formula:
$\frac{x-3}{6-3} = \frac{y-4}{-1-4} = \frac{z-(-7)}{1-(-7)}$
$\frac{x-3}{3} = \frac{y-4}{-5} = \frac{z+7}{8}$.
Thus,the correct option is $A$.

(B) Let $Q$ be the foot of the perpendicular drawn from the point $P(0,2,3)$ to the given line.
Any point on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}=\lambda$ is given by $Q(5\lambda-3, 2\lambda+1, 3\lambda-4)$.
The direction ratios of the line $PQ$ are $(5\lambda-3-0, 2\lambda+1-2, 3\lambda-4-3)$,which simplifies to $(5\lambda-3, 2\lambda-1, 3\lambda-7)$.
The direction ratios of the given line are $(5, 2, 3)$.
Since $PQ$ is perpendicular to the line,the dot product of their direction ratios 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 $\lambda = 1$ into the coordinates of $Q$:
$Q = (5(1)-3, 2(1)+1, 3(1)-4) = (2, 3, -1)$.
Thus,the correct option is $(B)$.

(C) First,we write the equations of the lines in symmetric form $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$.
For the first line $1+x = 2y = -12z$,we have $\frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12}$. Here,point $P_1 = (-1, 0, 0)$ and direction vector $\vec{b_1} = (1, 1/2, -1/12)$.
For the second line $x = y+2 = 6z-6$,we have $\frac{x}{1} = \frac{y+2}{1} = \frac{z-1}{1/6}$. Here,point $P_2 = (0, -2, 1)$ and direction vector $\vec{b_2} = (1, 1, 1/6)$.
The shortest distance $d$ is given by $d = \frac{|(\vec{P_2} - \vec{P_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{||\vec{b_1} \times \vec{b_2}||}$.
Vector $\vec{P_2} - \vec{P_1} = (0 - (-1), -2 - 0, 1 - 0) = (1, -2, 1)$.
Cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1/2 & -1/12 \\ 1 & 1 & 1/6 \end{vmatrix} = \hat{i}(\frac{1}{12} + \frac{1}{12}) - \hat{j}(\frac{1}{6} + \frac{1}{12}) + \hat{k}(1 - 1/2) = (1/6, -1/4, 1/2)$.
Magnitude $||\vec{b_1} \times \vec{b_2}|| = \sqrt{(1/6)^2 + (-1/4)^2 + (1/2)^2} = \sqrt{1/36 + 1/16 + 1/4} = \sqrt{\frac{4+9+36}{144}} = \sqrt{49/144} = 7/12$.
Dot product $(\vec{P_2} - \vec{P_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (1)(1/6) + (-2)(-1/4) + (1)(1/2) = 1/6 + 1/2 + 1/2 = 7/6$.
Therefore,$d = \frac{|7/6|}{7/12} = \frac{7}{6} \times \frac{12}{7} = 2$ units.

(B) The direction ratios of the first line are $\vec{b_1} = (4, 1, 8)$.
The direction ratios of the second line are $\vec{b_2} = (2, 2, 1)$.
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 $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$.
Substituting the values:
$\cos \theta = \left| \frac{4(2) + 1(2) + 8(1)}{\sqrt{4^2 + 1^2 + 8^2} \sqrt{2^2 + 2^2 + 1^2}} \right|$
$\cos \theta = \left| \frac{8 + 2 + 8}{\sqrt{16 + 1 + 64} \sqrt{4 + 4 + 1}} \right|$
$\cos \theta = \left| \frac{18}{\sqrt{81} \cdot \sqrt{9}} \right| = \frac{18}{9 \cdot 3} = \frac{18}{27} = \frac{2}{3}$.
Therefore,$\theta = \cos^{-1}\left(\frac{2}{3}\right)$.

(A) Let the direction ratios of the two lines be $\vec{b_1} = (1, 2, 2)$ and $\vec{b_2} = (2, 2, -1)$.
The formula for 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:
$\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$
Substituting the values:
$\cos \theta = \left| \frac{1(2) + 2(2) + 2(-1)}{\sqrt{1^2 + 2^2 + 2^2} \sqrt{2^2 + 2^2 + (-1)^2}} \right|$
$\cos \theta = \left| \frac{2 + 4 - 2}{\sqrt{1 + 4 + 4} \sqrt{4 + 4 + 1}} \right|$
$\cos \theta = \left| \frac{4}{\sqrt{9} \sqrt{9}} \right| = \frac{4}{3 \times 3} = \frac{4}{9}$
Therefore,$\theta = \cos ^{-1}\left(\frac{4}{9}\right)$.

(A) The direction vectors of the two lines are $\vec{b_1} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b_2} = 2\hat{i} + \hat{j} - \hat{k}$.
Let $\theta$ be the angle between the lines.
The formula for the angle between two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(2) + (1)(1) + (2)(-1) = 2 + 1 - 2 = 1$.
Calculating the magnitudes: $|\vec{b_1}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$ and $|\vec{b_2}| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{6}$.
Substituting these values into the formula: $\cos \theta = \frac{|1|}{\sqrt{6} \cdot \sqrt{6}} = \frac{1}{6}$.
Therefore,$\theta = \cos^{-1}\left(\frac{1}{6}\right)$.

(D) Let $\vec{a}$ and $\vec{b}$ be the direction vectors of the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ respectively.
$\vec{a} = 4\hat{i} + \hat{j} + 8\hat{k}$ and $\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}$.
The dot product is $\vec{a} \cdot \vec{b} = (4 \times 2) + (1 \times 2) + (8 \times 1) = 8 + 2 + 8 = 18$.
The magnitudes are $|\vec{a}| = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9$ and $|\vec{b}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Let $\theta$ be the angle between the lines. Then $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{18}{9 \times 3} = \frac{18}{27} = \frac{2}{3}$.
Therefore,$\theta = \cos^{-1}\left(\frac{2}{3}\right)$.

(B) The direction vectors of the two lines are $\vec{b_1} = \hat{\imath} - 2\hat{\jmath} - 2\hat{k}$ and $\vec{b_2} = 3\hat{\imath} + 2\hat{\jmath} - 6\hat{k}$.
The cosine of the angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
First,calculate the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(3) + (-2)(2) + (-2)(-6) = 3 - 4 + 12 = 11$.
Next,calculate the magnitudes: $|\vec{b_1}| = \sqrt{1^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$|\vec{b_2}| = \sqrt{3^2 + 2^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.
Therefore,$\cos \theta = \frac{|11|}{3 \times 7} = \frac{11}{21}$.

(D) The direction vector of the first line is $\vec{b_1} = \hat{i} - \hat{j} + \hat{k}$.
The direction vector of the second line is $\vec{b_2} = \hat{i} + 3\hat{j} + 2\hat{k}$.
The angle $\theta$ between the two lines is given by $\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$.
Calculating the dot product: $\vec{b_1} \cdot \vec{b_2} = (1)(1) + (-1)(3) + (1)(2) = 1 - 3 + 2 = 0$.
Since the dot product is $0$,$\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$.

(A) Let the direction ratios of the required line be $(a, b, c)$.
The direction ratios of the first line are $(1, 2, 4)$ and the second line are $(2, 2, 1)$.
Since the required line is perpendicular to both,its direction vector is the cross product of the direction vectors of the given lines:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-8) - \hat{j}(1-8) + \hat{k}(2-4) = -6\hat{i} + 7\hat{j} - 2\hat{k}$.
Thus,the direction ratios are proportional to $(-6, 7, -2)$ or $(6, -7, 2)$.
The line passes through $(1, 2, 3)$,so its equation is $\frac{x-1}{6} = \frac{y-2}{-7} = \frac{z-3}{2}$.
This can be rewritten as $\frac{x-1}{6} = \frac{2-y}{7} = \frac{z-3}{2}$.

(C) The vector equation of a line passing through points $A(\vec{a})$ and $B(\vec{b})$ is given by $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.
Here,$\vec{a} = 3\hat{i} + 4\hat{j} - 7\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 6\hat{k}$.
The direction vector $\vec{v} = \vec{b} - \vec{a} = (1-3)\hat{i} + (-1-4)\hat{j} + (6 - (-7))\hat{k} = -2\hat{i} - 5\hat{j} + 13\hat{k}$.
Thus,the parametric equations are $x = x_1 + v_x \lambda$,$y = y_1 + v_y \lambda$,$z = z_1 + v_z \lambda$.
Substituting the values: $x = 3 - 2\lambda$,$y = 4 - 5\lambda$,$z = -7 + 13\lambda$.

(D) Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Given the direction ratios of the first line are $(-3, 2k, 2)$ and the second line are $(3k, 1, -5)$.
Applying the condition for perpendicularity:
$(-3)(3k) + (2k)(1) + (2)(-5) = 0$
$-9k + 2k - 10 = 0$
$-7k - 10 = 0$
$-7k = 10$
$k = \frac{-10}{7}$

(A) The equation of a plane parallel to $2x + 3y - 4z = 17$ is of the form $2x + 3y - 4z = d$.
Since the plane passes through the point $(1, -1, 1)$,we substitute these coordinates into the equation:
$2(1) + 3(-1) - 4(1) = d$
$2 - 3 - 4 = d$
$d = -5$.
Thus,the Cartesian equation is $2x + 3y - 4z = -5$.
In vector form,this is $\bar{r} \cdot (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = -5$.

(D) The length of the perpendicular from a point with position vector $\bar{a}$ to the plane $\bar{r} \cdot \bar{n} = p$ is given by the formula $d = \frac{|\bar{a} \cdot \bar{n} - p|}{|\bar{n}|}$.
Here,the origin is the point,so $\bar{a} = 0\hat{i} + 0\hat{j} + 0\hat{k}$.
The plane equation is $\bar{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 14$,so $\bar{n} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $p = 14$.
Calculating the dot product: $\bar{a} \cdot \bar{n} = 0(1) + 0(-2) + 0(3) = 0$.
Calculating the magnitude of the normal vector: $|\bar{n}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
Substituting these values into the formula: $d = \frac{|0 - 14|}{\sqrt{14}} = \frac{14}{\sqrt{14}} = \sqrt{14}$ units.

(C) The intercept form of the plane $E_{1}$ with intercepts $a=1, b=-3, c=4$ is $\frac{x}{1} + \frac{y}{-3} + \frac{z}{4} = 1$.
Multiplying by $12$,we get $12x - 4y + 3z = 12$,or $12x - 4y + 3z - 12 = 0$.
The normal vector to this plane is $\vec{n} = 12\hat{i} - 4\hat{j} + 3\hat{k}$.
Since the required plane is parallel to $E_{1}$,its equation is of the form $12x - 4y + 3z + k = 0$.
Since it passes through $(2, 6, -8)$,we substitute these coordinates: $12(2) - 4(6) + 3(-8) + k = 0$.
$24 - 24 - 24 + k = 0 \Rightarrow k = 24$.
Thus,the equation is $12x - 4y + 3z + 24 = 0$.
Dividing by $12$,we get $x - \frac{y}{3} + \frac{z}{4} + 2 = 0$.

(C) The equation of a plane passing through the point $(2,3,1)$ is given by $a(x-2)+b(y-3)+c(z-1)=0$ $\ldots(1)$.
Since the point $(4,-5,3)$ lies on the plane,we substitute these coordinates into equation $(1)$:
$a(4-2)+b(-5-3)+c(3-1)=0$
$2a-8b+2c=0$
$a-4b+c=0$ $\ldots(2)$.
Since the plane is parallel to the $y$-axis,its normal vector is perpendicular to the $y$-axis vector $(0,1,0)$. Thus,$a(0)+b(1)+c(0)=0$,which implies $b=0$.
Substituting $b=0$ into equation $(2)$,we get $a+c=0$,or $a=-c$.
Substituting $a=-c$ and $b=0$ into equation $(1)$:
$-c(x-2)+0(y-3)+c(z-1)=0$
Dividing by $-c$ (assuming $c \neq 0$):
$(x-2)-(z-1)=0$
$x-z-1=0$
$x-z=1$.

(B) The direction ratios of the normal to the plane $x+y+3z-4=0$ are $(1, 1, 3)$.
Since the perpendicular line passes through the origin $(0, 0, 0)$,its equation is $\frac{x}{1} = \frac{y}{1} = \frac{z}{3} = K$.
Any point on this line is of the form $P(K, K, 3K)$.
Since this point $P$ lies on the plane,it must satisfy the plane equation:
$K + K + 3(3K) - 4 = 0$
$2K + 9K = 4$
$11K = 4 \Rightarrow K = \frac{4}{11}$.
Substituting $K$ back into the coordinates of $P$,we get $P = \left(\frac{4}{11}, \frac{4}{11}, \frac{12}{11}\right)$.

(D) The plane bisects the segment $AB$ perpendicularly,meaning it passes through the midpoint of $AB$ and the vector $\vec{AB}$ is normal to the plane.
Midpoint $M = \left(\frac{3+1}{2}, \frac{2+4}{2}, \frac{-1+3}{2}\right) = (2, 3, 1)$.
The normal vector $\vec{n}$ is the direction ratio of line $AB$: $\vec{n} = (1-3, 4-2, 3-(-1)) = (-2, 2, 4)$.
We can simplify the normal vector by dividing by $-2$,giving $\vec{n}' = (1, -1, -2)$.
The equation of the plane passing through $(x_0, y_0, z_0)$ with normal $(a, b, c)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$.
Substituting the values: $1(x-2) - 1(y-3) - 2(z-1) = 0$.
$x - 2 - y + 3 - 2z + 2 = 0$.
$x - y - 2z + 3 = 0$.

(C) The equation of a plane passing through the intersection of two planes $P_1: x+2y-3z+2=0$ and $P_2: 6x+y+z+1=0$ is given by $P_1 + \lambda P_2 = 0$.
$(x+2y-3z+2) + \lambda(6x+y+z+1) = 0$
$(1+6\lambda)x + (2+\lambda)y + (-3+\lambda)z + (2+\lambda) = 0$
This plane is parallel to the line $\frac{x-1}{1} = \frac{y+2}{1} = \frac{z-7}{-1}$.
The normal vector of the plane is $\vec{n} = (1+6\lambda, 2+\lambda, -3+\lambda)$ and the direction vector of the line is $\vec{v} = (1, 1, -1)$.
Since the plane is parallel to the line,the normal vector is perpendicular to the direction vector,so $\vec{n} \cdot \vec{v} = 0$.
$(1+6\lambda)(1) + (2+\lambda)(1) + (-3+\lambda)(-1) = 0$
$1 + 6\lambda + 2 + \lambda + 3 - \lambda = 0$
$6\lambda + 6 = 0 \Rightarrow \lambda = -1$.
Substituting $\lambda = -1$ into the equation:
$(1-6)x + (2-1)y + (-3-1)z + (2-1) = 0$
$-5x + y - 4z + 1 = 0$
$5x - y + 4z = 1$.

(D) The normal vector $\vec{n}$ of the plane is perpendicular to the direction vectors of both lines,$\vec{v}_1 = 2\hat{i} + 3\hat{j} + 6\hat{k}$ and $\vec{v}_2 = \hat{i} + \hat{j} - \hat{k}$.
Thus,$\vec{n} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{vmatrix} = \hat{i}(-3-6) - \hat{j}(-2-6) + \hat{k}(2-3) = -9\hat{i} + 8\hat{j} - \hat{k}$.
We can take the normal vector as $\vec{n} = 9\hat{i} - 8\hat{j} + \hat{k}$.
The plane passes through the point $(1, 3, 4)$ from the second line equation.
The equation of the plane is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,where $\vec{a} = \hat{i} + 3\hat{j} + 4\hat{k}$.
$(x\hat{i} + y\hat{j} + z\hat{k} - (\hat{i} + 3\hat{j} + 4\hat{k})) \cdot (9\hat{i} - 8\hat{j} + \hat{k}) = 0$.
$9(x-1) - 8(y-3) + 1(z-4) = 0$.
$9x - 9 - 8y + 24 + z - 4 = 0$.
$9x - 8y + z + 11 = 0$.

(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 the points $A(1, 1, \lambda)$ and $B(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$.
Distance from $A$: $d_1 = \frac{|3(1) + 4(1) - 12(\lambda) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|20 - 12\lambda|}{\sqrt{9 + 16 + 144}} = \frac{|20 - 12\lambda|}{13}$.
Distance from $B$: $d_2 = \frac{|3(-3) + 4(0) - 12(1) + 13|}{\sqrt{3^2 + 4^2 + (-12)^2}} = \frac{|-9 + 0 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$.
Since $d_1 = d_2$,we have $\frac{|20 - 12\lambda|}{13} = \frac{8}{13}$,which implies $|20 - 12\lambda| = 8$.
Case $1$: $20 - 12\lambda = 8 \Rightarrow 12\lambda = 12 \Rightarrow \lambda = 1$.
Case $2$: $20 - 12\lambda = -8 \Rightarrow 12\lambda = 28 \Rightarrow \lambda = \frac{28}{12} = \frac{7}{3}$.
Since $\lambda$ must be an integer,the value is $\lambda = 1$.

(D) The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by the formula:
$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$
Here,the point is $(2, -1, 0)$ and the plane is $2x + y + 2z + 8 = 0$.
Substituting the values into the formula:
$d = \frac{|2(2) + 1(-1) + 2(0) + 8|}{\sqrt{2^2 + 1^2 + 2^2}}$
$d = \frac{|4 - 1 + 0 + 8|}{\sqrt{4 + 1 + 4}}$
$d = \frac{|11|}{\sqrt{9}}$
$d = \frac{11}{3}$ units.

(B) Let the foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane be $P(3, 2, 1)$.
The normal vector $\vec{n}$ to the plane is the vector $\vec{OP} = 3\hat{i} + 2\hat{j} + \hat{k}$.
The equation of a plane passing through a point $P(x_1, y_1, z_1)$ with normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.
Substituting the values $a=3, b=2, c=1$ and $(x_1, y_1, z_1) = (3, 2, 1)$:
$3(x - 3) + 2(y - 2) + 1(z - 1) = 0$
$3x - 9 + 2y - 4 + z - 1 = 0$
$3x + 2y + z - 14 = 0$
$3x + 2y + z = 14$.
Thus,the correct option is $B$.

(C) The given equation of the plane is $2x + 3y + 5z = 1$.
To find the intercepts on the coordinate axes,we rewrite the equation in the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
This gives $\frac{x}{(1/2)} + \frac{y}{(1/3)} + \frac{z}{(1/5)} = 1$.
Thus,the intercepts on the $X, Y, Z$ axes are $a = 1/2$,$b = 1/3$,and $c = 1/5$ respectively.
The coordinates of the points $A, B, C$ are $A = (1/2, 0, 0)$,$B = (0, 1/3, 0)$,and $C = (0, 0, 1/5)$.
The centroid of $\triangle ABC$ is given by the formula $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Substituting the coordinates,we get $\left(\frac{1/2 + 0 + 0}{3}, \frac{0 + 1/3 + 0}{3}, \frac{0 + 0 + 1/5}{3}\right) = \left(\frac{1}{6}, \frac{1}{9}, \frac{1}{15}\right)$.
Therefore,the correct option is $C$.

(D) The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by the formula:
$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$
Given the point $(1, 2, -1)$ and the plane $x - 2y + 4z + 10 = 0$,we have $A=1, B=-2, C=4, D=10$ and $x_1=1, y_1=2, z_1=-1$.
Substituting these values into the formula:
$d = \frac{|1(1) - 2(2) + 4(-1) + 10|}{\sqrt{1^2 + (-2)^2 + 4^2}}$
$d = \frac{|1 - 4 - 4 + 10|}{\sqrt{1 + 4 + 16}}$
$d = \frac{|3|}{\sqrt{21}} = \frac{3}{\sqrt{21}} = \frac{3}{\sqrt{3 \times 7}} = \frac{\sqrt{3}}{\sqrt{7}} = \sqrt{\frac{3}{7}}$ units.
Thus,the correct option is $D$.

(A) The direction vectors of the two given lines are $\vec{b_1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and $\vec{b_2} = -3 \hat{i} + 2 \hat{j} + 5 \hat{k}$.
Since the required line is perpendicular to both,its direction vector $\vec{v}$ is given by the cross product $\vec{b_1} \times \vec{b_2}$.
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & 2 & 5 \end{vmatrix} = \hat{i}(10 - 6) - \hat{j}(5 - (-9)) + \hat{k}(2 - (-6)) = 4 \hat{i} - 14 \hat{j} + 8 \hat{k}$.
The line passes through $(1, 2, 3)$,so its vector equation is $\bar{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + \lambda(4 \hat{i} - 14 \hat{j} + 8 \hat{k})$.

(A) The equation of any plane parallel to $x+2y+2z+8=0$ is of the form $x+2y+2z+\lambda=0$.
Given that the distance of this plane from the point $(1,1,2)$ is $2$ units.
The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax+By+Cz+D=0$ is given by $d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$.
Substituting the values,we get:
$2 = \frac{|1(1)+2(1)+2(2)+\lambda|}{\sqrt{1^2+2^2+2^2}}$
$2 = \frac{|1+2+4+\lambda|}{\sqrt{1+4+4}}$
$2 = \frac{|7+\lambda|}{\sqrt{9}}$
$2 = \frac{|7+\lambda|}{3}$
$|7+\lambda| = 6$
This implies $7+\lambda = 6$ or $7+\lambda = -6$.
Case $1$: $7+\lambda = 6 \Rightarrow \lambda = -1$.
Case $2$: $7+\lambda = -6 \Rightarrow \lambda = -13$.
Thus,the equations of the planes are $x+2y+2z-1=0$ and $x+2y+2z-13=0$.

(A) The equation of a plane passing through $(1, -1, 2)$ is given by $a(x - 1) + b(y + 1) + c(z - 2) = 0$.
Since the plane is perpendicular to the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$,its normal vector $\vec{n} = (a, b, c)$ must be perpendicular to the normal vectors of the given planes,$\vec{n_1} = (2, 3, -2)$ and $\vec{n_2} = (1, 2, -3)$.
Thus,$\vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -2 \\ 1 & 2 & -3 \end{vmatrix} = \hat{i}(-9 - (-4)) - \hat{j}(-6 - (-2)) + \hat{k}(4 - 3) = -5\hat{i} + 4\hat{j} + \hat{k}$.
So,$a = -5, b = 4, c = 1$.
Substituting these values into the plane equation: $-5(x - 1) + 4(y + 1) + 1(z - 2) = 0$.
$-5x + 5 + 4y + 4 + z - 2 = 0 \Rightarrow -5x + 4y + z + 7 = 0 \Rightarrow 5x - 4y - z = 7$.
In vector form,this is $\bar{r} \cdot (5\hat{i} - 4\hat{j} - \hat{k}) = 7$.

(C) The given plane is $2x + 3y - 4z = 17$. The normal vector to this plane is $\overline{n} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$.
Since the required plane is parallel to the given plane,it will have the same normal vector $\overline{n} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$.
The equation of a plane passing through point $\overline{a} = \hat{i} - \hat{j} + \hat{k}$ and having normal $\overline{n}$ is given by $\overline{r} \cdot \overline{n} = \overline{a} \cdot \overline{n}$.
Calculating $\overline{a} \cdot \overline{n} = (1)(2) + (-1)(3) + (1)(-4) = 2 - 3 - 4 = -5$.
Thus,the equation of the plane is $\overline{r} \cdot (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = -5$.

(B) The equation of the plane is given by $4x - 3y + 12z = 15$.
Comparing this with the general form $ax + by + cz = d$,we get the normal vector $\vec{n} = 4\hat{i} - 3\hat{j} + 12\hat{k}$.
The magnitude of the normal vector is $|\vec{n}| = \sqrt{4^2 + (-3)^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$.
The unit vector perpendicular to the plane is given by $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{4\hat{i} - 3\hat{j} + 12\hat{k}}{13}$.

(D) The direction ratios of the given lines are $\vec{b_1} = (1, 4, 7)$ and $\vec{b_2} = (3, 5, 7)$.
Since the plane contains the first line and is parallel to the second,the normal vector $\vec{n}$ to the plane is perpendicular to both $\vec{b_1}$ and $\vec{b_2}$.
$\vec{n} = \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & 7 \\ 3 & 5 & 7 \end{vmatrix}$
$= \hat{i}(28-35) - \hat{j}(7-21) + \hat{k}(5-12) = -7\hat{i} + 14\hat{j} - 7\hat{k} = -7(\hat{i} - 2\hat{j} + \hat{k})$.
Thus,the equation of the plane is of the form $x - 2y + z = d$.
Since the plane contains the line,it must pass through the point $(2, 4, 6)$.
Substituting the point into the equation: $2 - 2(4) + 6 = 2 - 8 + 6 = 0$.
Therefore,the equation of the plane is $x - 2y + z = 0$.

(A) Since the points $O(0, 0, 0)$,$A(1, 2, 3)$,$B(2, 3, 4)$,and $P(x, y, z)$ are coplanar,the scalar triple product of the vectors $\vec{OA}$,$\vec{OB}$,and $\vec{OP}$ must be zero.
The condition for coplanarity is given by the determinant:
$\begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ x & y & z \end{vmatrix} = 0$
Expanding the determinant along the third row:
$x(2 \times 4 - 3 \times 3) - y(1 \times 4 - 3 \times 2) + z(1 \times 3 - 2 \times 2) = 0$
$x(8-9) - y(4-6) + z(3-4) = 0$
$-x - y(-2) + z(-1) = 0$
$-x + 2y - z = 0$
Multiplying by $-1$,we get:
$x - 2y + z = 0$

(D) Let the given lines be $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=\mu$.
Any point on the first line is $(2\lambda+1, 3\lambda-1, 4\lambda+1)$ and any point on the second line is $(\mu+3, 2\mu+k, \mu)$.
Since the lines intersect,there must exist some $\lambda$ and $\mu$ such that the coordinates are equal:
$2\lambda+1 = \mu+3 \implies 2\lambda - \mu = 2$ .... $(1)$
$3\lambda-1 = 2\mu+k \implies 3\lambda - 2\mu = k+1$ .... $(2)$
$4\lambda+1 = \mu \implies 4\lambda - \mu = -1$ .... $(3)$
Subtracting $(1)$ from $(3)$,we get $(4\lambda - \mu) - (2\lambda - \mu) = -1 - 2$,which gives $2\lambda = -3$,so $\lambda = \frac{-3}{2}$.
Substituting $\lambda = \frac{-3}{2}$ into $(3)$,we get $\mu = 4(\frac{-3}{2}) + 1 = -6 + 1 = -5$.
Now,substitute $\lambda = \frac{-3}{2}$ and $\mu = -5$ into $(2)$:
$3(\frac{-3}{2}) - 2(-5) = k+1$
$\frac{-9}{2} + 10 = k+1$
$\frac{-9+20}{2} = k+1$
$\frac{11}{2} = k+1$
$k = \frac{11}{2} - 1 = \frac{9}{2}$.

(A) The angle $\theta$ between a line with direction ratios $(a, b, c)$ and a plane with normal vector $\vec{n} = (a_1, b_1, c_1)$ is given by the formula:
$\sin \theta = \frac{|a a_1 + b b_1 + c c_1|}{\sqrt{a^2 + b^2 + c^2} \sqrt{a_1^2 + b_1^2 + c_1^2}}$
Here,the direction ratios of the line are $(a, b, c) = (2, 1, 2)$ and the normal vector to the plane is $\vec{n} = (6, -2, -3)$.
The magnitude of the direction vector of the line is $\sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.
The magnitude of the normal vector to the plane is $\sqrt{6^2 + (-2)^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.
Substituting these values into the formula:
$\sin \theta = \frac{|(2)(6) + (1)(-2) + (2)(-3)|}{3 \times 7} = \frac{|12 - 2 - 6|}{21} = \frac{4}{21}$.
Therefore,$\theta = \sin ^{-1}\left(\frac{4}{21}\right)$.

(D) The direction ratios of the line $OP$ are $(1, \sqrt{2}, 1)$.
Let the direction cosines of the line $OP$ be $(l, m, n)$.
The magnitude of vector $\vec{OP}$ is $\sqrt{1^2 + (\sqrt{2})^2 + 1^2} = \sqrt{1 + 2 + 1} = \sqrt{4} = 2$.
Thus,the direction cosines are $l = \frac{1}{2}$,$m = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$,and $n = \frac{1}{2}$.
The angle $\alpha$ made by a line with direction cosines $(l, m, n)$ with the $YOZ$ plane is given by $\sin \alpha = |l| = \frac{1}{2}$,so $\alpha = 30^{\circ}$.
The angle $\beta$ made with the $ZOX$ plane is given by $\sin \beta = |m| = \frac{1}{\sqrt{2}}$,so $\beta = 45^{\circ}$.
The angle $\gamma$ made with the $XOY$ plane is given by $\sin \gamma = |n| = \frac{1}{2}$,so $\gamma = 30^{\circ}$.
Wait,re-evaluating the question order: $XOY$ plane (involves $z$),$YOZ$ plane (involves $x$),$ZOX$ plane (involves $y$).
Angle with $XOY$ plane: $\sin \theta_1 = |n| = \frac{1}{2} \implies \theta_1 = 30^{\circ}$.
Angle with $YOZ$ plane: $\sin \theta_2 = |l| = \frac{1}{2} \implies \theta_2 = 30^{\circ}$.
Angle with $ZOX$ plane: $\sin \theta_3 = |m| = \frac{1}{\sqrt{2}} \implies \theta_3 = 45^{\circ}$.
Given the options,there might be a convention difference. Let's re-calculate: The angle $\theta$ between a line with direction cosines $(l, m, n)$ and a plane with normal $(a, b, c)$ is $\sin \theta = |al + bm + cn|$.
For $XOY$ ($z=0$,normal $(0,0,1)$): $\sin \theta_1 = |0(1/2) + 0(1/\sqrt{2}) + 1(1/2)| = 1/2 \implies 30^{\circ}$.
For $YOZ$ ($x=0$,normal $(1,0,0)$): $\sin \theta_2 = |1(1/2) + 0 + 0| = 1/2 \implies 30^{\circ}$.
For $ZOX$ ($y=0$,normal $(0,1,0)$): $\sin \theta_3 = |0 + 1(1/\sqrt{2}) + 0| = 1/\sqrt{2} \implies 45^{\circ}$.
Correct answer is $30^{\circ}, 30^{\circ}, 45^{\circ}$.

(A) The angle $\theta$ between a line with direction vector $\bar{b}$ and a plane with normal vector $\bar{n}$ is given by $\sin \theta = \frac{|\bar{b} \cdot \bar{n}|}{|\bar{b}| |\bar{n}|}$.
Given the line $\bar{r}=(\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}+\hat{j}+\hat{k})$,the direction vector is $\bar{b} = \hat{i}+\hat{j}+\hat{k}$.
Given the plane $\bar{r} \cdot(2\hat{i}-\hat{j}+\hat{k})=5$,the normal vector is $\bar{n} = 2\hat{i}-\hat{j}+\hat{k}$.
Calculate the magnitudes: $|\bar{b}| = \sqrt{1^2+1^2+1^2} = \sqrt{3}$ and $|\bar{n}| = \sqrt{2^2+(-1)^2+1^2} = \sqrt{4+1+1} = \sqrt{6}$.
Calculate the dot product: $\bar{b} \cdot \bar{n} = (1)(2) + (1)(-1) + (1)(1) = 2 - 1 + 1 = 2$.
Substitute into the formula: $\sin \theta = \frac{|2|}{\sqrt{3} \cdot \sqrt{6}} = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}$.
Therefore,$\theta = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right)$.

(A) Let the line be $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2} = \lambda$.
Any point on the line is given by $(x, y, z) = (\lambda+3, 2\lambda+4, 2\lambda+5)$.
Since this point lies on the plane $x+y+z=2$,we substitute the coordinates into the plane equation:
$(\lambda+3) + (2\lambda+4) + (2\lambda+5) = 2$.
$5\lambda + 12 = 2$.
$5\lambda = -10$,which gives $\lambda = -2$.
Substituting $\lambda = -2$ back into the point coordinates,we get the point of intersection:
$x = -2+3 = 1$,$y = 2(-2)+4 = 0$,$z = 2(-2)+5 = 1$.
The point of intersection is $(1, 0, 1)$.
The distance between $(3, 4, 5)$ and $(1, 0, 1)$ is calculated using the distance formula:
$d = \sqrt{(3-1)^2 + (4-0)^2 + (5-1)^2} = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$ units.

(C) line given by $\bar{r} = \bar{a} + \lambda \bar{b}$ is parallel to a plane $\bar{r} \cdot \bar{n} = d$ if and only if the direction vector of the line $\bar{b}$ is perpendicular to the normal vector of the plane $\bar{n}$.
This implies that the dot product of the vectors $\bar{b}$ and $\bar{n}$ must be zero,i.e.,$\bar{b} \cdot \bar{n} = 0$.
Given $\bar{b} = 2 \hat{i} + \hat{j} + 2 \hat{k}$ and $\bar{n} = 3 \hat{i} - 2 \hat{j} - m \hat{k}$.
Calculating the dot product: $(2 \hat{i} + \hat{j} + 2 \hat{k}) \cdot (3 \hat{i} - 2 \hat{j} - m \hat{k}) = 0$.
$(2)(3) + (1)(-2) + (2)(-m) = 0$.
$6 - 2 - 2m = 0$.
$4 - 2m = 0$.
$2m = 4$.
$m = 2$.

(D) Let $P = (7, 5, 2)$.
The equation of the line passing through $P$ and parallel to the given line $\frac{x-1}{3} = \frac{y-2}{6} = \frac{z+1}{2}$ is given by $\frac{x-7}{3} = \frac{y-5}{6} = \frac{z-2}{2} = r$.
Any point $Q$ on this line can be represented as $(3r+7, 6r+5, 2r+2)$.
Since $Q$ lies on the plane $3x+4y+z-8=0$,we substitute the coordinates of $Q$ into the plane equation:
$3(3r+7) + 4(6r+5) + (2r+2) - 8 = 0$.
Expanding the terms:
$9r + 21 + 24r + 20 + 2r + 2 - 8 = 0$.
Combining like terms:
$35r + 35 = 0 \Rightarrow 35r = -35 \Rightarrow r = -1$.
Substituting $r = -1$ back into the coordinates of $Q$:
$Q = (3(-1)+7, 6(-1)+5, 2(-1)+2) = (4, -1, 0)$.
The distance $PQ$ is calculated using the distance formula:
$PQ = \sqrt{(7-4)^2 + (5-(-1))^2 + (2-0)^2} = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$ units.

(C) The angle $\theta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin \theta = \left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|}\right|$.
Here,the direction vector of the line is $\vec{b} = 3\hat{i} + \hat{j}$ and the normal vector to the plane is $\vec{n} = \hat{i} + 2\hat{j} + 3\hat{k}$.
The dot product is $\vec{b} \cdot \vec{n} = (3)(1) + (1)(2) + (0)(3) = 3 + 2 + 0 = 5$.
The magnitudes are $|\vec{b}| = \sqrt{3^2 + 1^2} = \sqrt{10}$ and $|\vec{n}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$.
Substituting these values into the formula: $\sin \theta = \frac{5}{\sqrt{10} \cdot \sqrt{14}} = \frac{5}{\sqrt{140}} = \frac{5}{2\sqrt{35}} = \frac{\sqrt{5} \cdot \sqrt{5}}{2\sqrt{7} \cdot \sqrt{5}} = \frac{\sqrt{5}}{2\sqrt{7}}$.
Thus,$\theta = \sin^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right)$.

(C) The equation of the line is $\bar{r} = \bar{a} + \lambda \bar{b}$,where $\bar{b} = 2 \hat{\imath} + \hat{\jmath} + 2 \hat{k}$.
The equation of the plane is $\bar{r} \cdot \bar{n} = d$,where $\bar{n} = 3 \hat{\imath} - 2 \hat{\jmath} + m \hat{k}$.
Since the line is parallel to the plane,the direction vector of the line $\bar{b}$ must be perpendicular to the normal vector of the plane $\bar{n}$.
Therefore,$\bar{b} \cdot \bar{n} = 0$.
$(2 \hat{\imath} + \hat{\jmath} + 2 \hat{k}) \cdot (3 \hat{\imath} - 2 \hat{\jmath} + m \hat{k}) = 0$.
$(2)(3) + (1)(-2) + (2)(m) = 0$.
$6 - 2 + 2m = 0$.
$4 + 2m = 0$.
$2m = -4$.
$m = -2$.

(A) Let the given line be $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}=\lambda$.
Any point $P$ on the line is given by $P = (2\lambda+1, -3\lambda+2, 4\lambda-3)$.
Since this point $P$ lies on the plane $2x+4y-z=1$,we substitute the coordinates of $P$ into the plane equation:
$2(2\lambda+1) + 4(-3\lambda+2) - (4\lambda-3) = 1$.
Expanding the terms:
$4\lambda + 2 - 12\lambda + 8 - 4\lambda + 3 = 1$.
Combining the $\lambda$ terms and constants:
$-12\lambda + 13 = 1$.
$-12\lambda = -12$,which gives $\lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates of $P$:
$x = 2(1)+1 = 3$,
$y = -3(1)+2 = -1$,
$z = 4(1)-3 = 1$.
Thus,the point of intersection is $(3, -1, 1)$.

(C) The angle $\theta$ between the line $\bar{r}=\bar{a}+\lambda\bar{b}$ and the plane $\bar{r} \cdot \bar{n}=p$ is given by $\sin \theta = \frac{|\bar{b} \cdot \bar{n}|}{|\bar{b}| |\bar{n}|}$.
Here,$\bar{b} = \hat{i} - \hat{j} + \hat{k}$ and $\bar{n} = 2\hat{i} - \hat{j} + \hat{k}$.
Calculating the dot product: $\bar{b} \cdot \bar{n} = (1)(2) + (-1)(-1) + (1)(1) = 2 + 1 + 1 = 4$.
Calculating the magnitudes: $|\bar{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$ and $|\bar{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
Substituting these values into the formula: $\sin \theta = \frac{4}{\sqrt{3} \cdot \sqrt{6}} = \frac{4}{\sqrt{18}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}$.
Therefore,$\theta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$.

(D) The equation of the line is given by $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$.
Any point on this line can be represented in Cartesian coordinates as $(x, y, z) = (2+\lambda, 1-2\lambda, -4+2\lambda)$.
The $XOY$-plane is defined by the equation $z=0$.
Since the point of intersection lies on the $XOY$-plane,we set the $z$-coordinate to zero:
$-4+2\lambda = 0 \Rightarrow 2\lambda = 4 \Rightarrow \lambda = 2$.
Substituting $\lambda = 2$ back into the coordinates:
$x = 2+2 = 4$
$y = 1-2(2) = 1-4 = -3$
$z = -4+2(2) = 0$.
Thus,the point of intersection is $(4, -3, 0)$,and its position vector is $4 \hat{i}-3 \hat{j}$.

(B) Let $\cos ^{-1}\left(\frac{1}{3}\right)=\theta$,then $\cos \theta=\frac{1}{3}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\sin \theta = \sqrt{1-\left(\frac{1}{3}\right)^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.
Thus,the given equation is $\sin (x+y)+\cos (x+y)=\frac{2\sqrt{2}}{3}$.
Since $\frac{2\sqrt{2}}{3} \approx 0.9428$,and the maximum value of $\sin(x+y)+\cos(x+y)$ is $\sqrt{2} \approx 1.414$,this equation represents a constant value for the expression $(x+y)$.
Let $x+y = C$,where $C$ is a constant.
Differentiating both sides with respect to $x$:
$\frac{d}{dx}(x+y) = \frac{d}{dx}(C)$
$1 + \frac{dy}{dx} = 0$
$\frac{dy}{dx} = -1$.