Line Questions in English

(A) The direction ratios of the first line are $a_1 = 3, b_1 = 5, c_1 = 4$.
The direction ratios of the second line are $a_2 = 1, b_2 = 1, c_2 = 2$.
If $\theta$ is the angle between them,then the formula is:
$\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$
Substituting the values:
$\cos \theta = \left| \frac{(3)(1) + (5)(1) + (4)(2)}{\sqrt{3^2 + 5^2 + 4^2} \sqrt{1^2 + 1^2 + 2^2}} \right|$
$\cos \theta = \left| \frac{3 + 5 + 8}{\sqrt{9 + 25 + 16} \sqrt{1 + 1 + 4}} \right|$
$\cos \theta = \frac{16}{\sqrt{50} \sqrt{6}} = \frac{16}{5 \sqrt{2} \sqrt{6}} = \frac{16}{5 \sqrt{12}} = \frac{16}{5 \times 2 \sqrt{3}} = \frac{8}{5 \sqrt{3}}$
Rationalizing the denominator:
$\cos \theta = \frac{8 \sqrt{3}}{5 \times 3} = \frac{8 \sqrt{3}}{15}$
Therefore,$\theta = \cos ^{-1}\left(\frac{8 \sqrt{3}}{15}\right)$.

(A) Comparing $(1)$ and $(2)$ with $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$ respectively,
we get $\vec{a}_{1}=\hat{i}+\hat{j}, \quad \vec{b}_{1}=2 \hat{i}-\hat{j}+\hat{k}$
$\vec{a}_{2}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}_{2}=3 \hat{i}-5 \hat{j}+2 \hat{k}$
Therefore,$\vec{a}_{2}-\vec{a}_{1}=(2-1)\hat{i}+(1-1)\hat{j}+(-1-0)\hat{k} = \hat{i}-\hat{k}$
Now,$\vec{b}_{1} \times \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} = \hat{i}(-2+5) - \hat{j}(4-3) + \hat{k}(-10+3) = 3\hat{i} - \hat{j} - 7\hat{k}$
Magnitude $|\vec{b}_{1} \times \vec{b}_{2}| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9+1+49} = \sqrt{59}$
The shortest distance $d$ is given by $d = \left| \frac{(\vec{b}_{1} \times \vec{b}_{2}) \cdot (\vec{a}_{2}-\vec{a}_{1})}{|\vec{b}_{1} \times \vec{b}_{2}|} \right|$
$d = \left| \frac{(3\hat{i} - \hat{j} - 7\hat{k}) \cdot (\hat{i} - \hat{k})}{\sqrt{59}} \right| = \left| \frac{3(1) + (-1)(0) + (-7)(-1)}{\sqrt{59}} \right| = \left| \frac{3+0+7}{\sqrt{59}} \right| = \frac{10}{\sqrt{59}}$

(N/A) Let $AB$ be the line joining the points $(1, -1, 2)$ and $(3, 4, -2)$,and $CD$ be the line through the points $(0, 3, 2)$ and $(3, 5, 6)$.
The direction ratios $(a_1, b_1, c_1)$ of line $AB$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1) = (3-1, 4-(-1), -2-2) = (2, 5, -4)$.
The direction ratios $(a_2, b_2, c_2)$ of line $CD$ are given by $(x_4-x_3, y_4-y_3, z_4-z_3) = (3-0, 5-3, 6-2) = (3, 2, 4)$.
Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Calculating the sum of products of direction ratios:
$a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(3) + (5)(2) + (-4)(4)$
$= 6 + 10 - 16$
$= 16 - 16$
$= 0$.
Since the sum is $0$,the line $AB$ is perpendicular to the line $CD$.

(A) Let $AB$ be the line passing through the points $A(4,7,8)$ and $B(2,3,4)$.
Let $CD$ be the line passing through the points $C(-1,-2,1)$ and $D(1,2,5)$.
The direction ratios $(a_1, b_1, c_1)$ of line $AB$ are given by $(x_2-x_1, y_2-y_1, z_2-z_1)$:
$a_1 = 2-4 = -2$
$b_1 = 3-7 = -4$
$c_1 = 4-8 = -4$
The direction ratios $(a_2, b_2, c_2)$ of line $CD$ are given by $(x_4-x_3, y_4-y_3, z_4-z_3)$:
$a_2 = 1-(-1) = 2$
$b_2 = 2-(-2) = 4$
$c_2 = 5-1 = 4$
Two lines are parallel if their direction ratios are proportional,i.e.,$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Calculating the ratios:
$\frac{a_1}{a_2} = \frac{-2}{2} = -1$
$\frac{b_1}{b_2} = \frac{-4}{4} = -1$
$\frac{c_1}{c_2} = \frac{-4}{4} = -1$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = -1$,the direction ratios are proportional.
Therefore,the line $AB$ is parallel to the line $CD$.

(A) The equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$.
Given the point $(1, 2, 3)$,the position vector is $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$.
Given the parallel vector $\vec{b} = 3 \hat{i} + 2 \hat{j} - 2 \hat{k}$.
Substituting these into the formula,we get $\vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$.

(N/A) The line passes through the point with position vector $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$.
The direction of the line is given by the vector $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$.
The vector equation of a line passing through a point $\vec{a}$ and parallel to a vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda\vec{b}$,where $\lambda$ is a scalar.
Substituting the values,we get the vector equation:
$\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})$.
For the Cartesian form,let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.
Then $x\hat{i} + y\hat{j} + z\hat{k} = (2 + \lambda)\hat{i} + (-1 + 2\lambda)\hat{j} + (4 - \lambda)\hat{k}$.
Comparing the components,we have:
$x = 2 + \lambda \Rightarrow \lambda = x - 2$
$y = -1 + 2\lambda \Rightarrow \lambda = \frac{y + 1}{2}$
$z = 4 - \lambda \Rightarrow \lambda = 4 - z = \frac{z - 4}{-1}$
Equating the values of $\lambda$,we get the Cartesian equation:
$\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}$.

(A) The given line is $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.
The direction ratios of this line are $a = 3, b = 5, c = 6$.
Since the required line is parallel to the given line,its direction ratios will also be proportional to $3, 5, 6$.
The line passes through the point $(x_1, y_1, z_1) = (-2, 4, -5)$.
The cartesian equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $a, b, c$ is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Substituting the values,we get $\frac{x - (-2)}{3} = \frac{y - 4}{5} = \frac{z - (-5)}{6}$.
Thus,the required equation is $\frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6}$.

(A) The Cartesian equation of the line is given by $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} .$
The general form of the Cartesian equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} .$
Comparing the given equation with the general form,we get the point $(x_1, y_1, z_1) = (5, -4, 6)$ and direction ratios $(a, b, c) = (3, 7, 2) .$
The position vector of the point is $\vec{a} = 5 \hat{i} - 4 \hat{j} + 6 \hat{k} .$
The vector parallel to the line is $\vec{b} = 3 \hat{i} + 7 \hat{j} + 2 \hat{k} .$
The vector equation of a line is given by $\vec{r} = \vec{a} + \lambda \vec{b} .$
Substituting the values,we get $\vec{r} = (5 \hat{i} - 4 \hat{j} + 6 \hat{k}) + \lambda(3 \hat{i} + 7 \hat{j} + 2 \hat{k}) ,$ where $\lambda$ is a scalar.

(A) The required line passes through the origin $(0, 0, 0)$ and the point $(5, -2, 3).$
The position vector of the origin is $\vec{a} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0}.$
The direction ratios of the line passing through $(0, 0, 0)$ and $(5, -2, 3)$ are $(5-0, -2-0, 3-0) = (5, -2, 3).$
Thus,the line is parallel to the vector $\vec{b} = 5\hat{i} - 2\hat{j} + 3\hat{k}.$
The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r} = \vec{a} + \lambda\vec{b}.$
Substituting the values,we get $\vec{r} = \vec{0} + \lambda(5\hat{i} - 2\hat{j} + 3\hat{k}) = \lambda(5\hat{i} - 2\hat{j} + 3\hat{k}),$ where $\lambda \in R.$
The Cartesian equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}.$
Substituting $(x_1, y_1, z_1) = (0, 0, 0)$ and $(a, b, c) = (5, -2, 3),$ we get $\frac{x-0}{5} = \frac{y-0}{-2} = \frac{z-0}{3}.$
Therefore,the Cartesian equation is $\frac{x}{5} = \frac{y}{-2} = \frac{z}{3}.$

(D) Let the line passing through the points $P(3,-2,-5)$ and $Q(3,-2,6)$ be $PQ.$
Since the line passes through $P(3,-2,-5),$ its position vector $\vec{a}$ is given by $\vec{a} = 3\hat{i} - 2\hat{j} - 5\hat{k}.$
The direction ratios of the line $PQ$ are $(3-3, -2-(-2), 6-(-5)) = (0, 0, 11).$
The vector $\vec{b}$ in the direction of the line is $\vec{b} = 0\hat{i} + 0\hat{j} + 11\hat{k} = 11\hat{k}.$
The vector equation of the line is $\vec{r} = \vec{a} + \lambda\vec{b},$ where $\lambda \in R.$
Substituting the values,we get $\vec{r} = (3\hat{i} - 2\hat{j} - 5\hat{k}) + \lambda(11\hat{k}).$
The Cartesian equation of a line passing through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$ is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}.$
Substituting the points and direction ratios,we get $\frac{x-3}{0} = \frac{y+2}{0} = \frac{z+5}{11}.$

(A) Let $\bar{b}_{1}$ and $\bar{b}_{2}$ be the vectors parallel to the given lines.
The direction ratios of the first line are $(2, 5, -3)$,so $\bar{b}_{1} = 2 \hat{i} + 5 \hat{j} - 3 \hat{k}$.
The direction ratios of the second line are $(-1, 8, 4)$,so $\bar{b}_{2} = -\hat{i} + 8 \hat{j} + 4 \hat{k}$.
Calculate the magnitudes:
$|\bar{b}_{1}| = \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$.
$|\bar{b}_{2}| = \sqrt{(-1)^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9$.
Calculate the dot product:
$\bar{b}_{1} \cdot \bar{b}_{2} = (2)(-1) + (5)(8) + (-3)(4) = -2 + 40 - 12 = 26$.
The angle $Q$ between the lines is given by:
$\cos Q = \frac{|\bar{b}_{1} \cdot \bar{b}_{2}|}{|\bar{b}_{1}| |\bar{b}_{2}|} = \frac{|26|}{9 \sqrt{38}} = \frac{26}{9 \sqrt{38}}$.
Therefore,$Q = \cos ^{-1}\left(\frac{26}{9 \sqrt{38}}\right)$.

(B) Let $\vec{b}_{1}$ and $\vec{b}_{2}$ be the vectors parallel to the given pair of lines.
The direction ratios of the first line are $(2, 2, 1)$,so $\vec{b}_{1} = 2\hat{i} + 2\hat{j} + \hat{k}$.
The direction ratios of the second line are $(4, 1, 8)$,so $\vec{b}_{2} = 4\hat{i} + \hat{j} + 8\hat{k}$.
Calculate the magnitudes:
$|\vec{b}_{1}| = \sqrt{2^{2} + 2^{2} + 1^{2}} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
$|\vec{b}_{2}| = \sqrt{4^{2} + 1^{2} + 8^{2}} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9$.
Calculate the dot product:
$\vec{b}_{1} \cdot \vec{b}_{2} = (2)(4) + (2)(1) + (1)(8) = 8 + 2 + 8 = 18$.
If $\theta$ is the angle between the lines,then $\cos \theta = \frac{|\vec{b}_{1} \cdot \vec{b}_{2}|}{|\vec{b}_{1}| |\vec{b}_{2}|}$.
$\cos \theta = \frac{18}{3 \times 9} = \frac{18}{27} = \frac{2}{3}$.
Therefore,$\theta = \cos^{-1}\left(\frac{2}{3}\right)$.

(A) The given equations can be rewritten in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ as follows:
For the first line: $\frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2}$.
For the second line: $\frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$.
The direction ratios of the lines are $a_1 = -3, b_1 = \frac{2p}{7}, c_1 = 2$ and $a_2 = \frac{-3p}{7}, b_2 = 1, c_2 = -5$.
Two lines are perpendicular if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
Substituting the values: $(-3)\left(\frac{-3p}{7}\right) + \left(\frac{2p}{7}\right)(1) + (2)(-5) = 0$.
$\frac{9p}{7} + \frac{2p}{7} - 10 = 0$.
$\frac{11p}{7} = 10$.
$11p = 70$.
$p = \frac{70}{11}$.

(A) The equations of the given lines are $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$.
The direction ratios of the first line are $a_{1}=7, b_{1}=-5, c_{1}=1$.
The direction ratios of the second line are $a_{2}=1, b_{2}=2, c_{2}=3$.
Two lines with direction ratios $(a_{1}, b_{1}, c_{1})$ and $(a_{2}, b_{2}, c_{2})$ are perpendicular to each other if and only if $a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0$.
Substituting the values,we get:
$a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = (7 \times 1) + (-5 \times 2) + (1 \times 3)$
$= 7 - 10 + 3$
$= 0$.
Since the sum of the products of the corresponding direction ratios is $0$,the given lines are perpendicular to each other.

(A) The equations of the given lines are $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$.
Comparing with the given lines:
$\vec{a}_{1}=\hat{i}+2 \hat{j}+\hat{k}$,$\vec{b}_{1}=\hat{i}-\hat{j}+\hat{k}$
$\vec{a}_{2}=2 \hat{i}-\hat{j}-\hat{k}$,$\vec{b}_{2}=2 \hat{i}+\hat{j}+2 \hat{k}$
The shortest distance $d$ is given by $d=\left| \frac{(\vec{b}_{1} \times \vec{b}_{2}) \cdot (\vec{a}_{2}-\vec{a}_{1})}{|\vec{b}_{1} \times \vec{b}_{2}|} \right|$.
First,calculate $\vec{a}_{2}-\vec{a}_{1} = (2\hat{i}-\hat{j}-\hat{k}) - (\hat{i}+2\hat{j}+\hat{k}) = \hat{i}-3\hat{j}-2\hat{k}$.
Next,calculate the cross product $\vec{b}_{1} \times \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = \hat{i}(-2-1) - \hat{j}(2-2) + \hat{k}(1+2) = -3\hat{i}+3\hat{k}$.
The magnitude is $|\vec{b}_{1} \times \vec{b}_{2}| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$.
Now,the dot product $(\vec{b}_{1} \times \vec{b}_{2}) \cdot (\vec{a}_{2}-\vec{a}_{1}) = (-3\hat{i}+3\hat{k}) \cdot (\hat{i}-3\hat{j}-2\hat{k}) = (-3)(1) + (0)(-3) + (3)(-2) = -3 - 6 = -9$.
Finally,$d = \left| \frac{-9}{3\sqrt{2}} \right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$ units.

(A) The given lines are $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$.
The shortest distance $d$ between two lines $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ and $\frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}$ is given by:
$d = \frac{|\det(A)|}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^{2} + (c_{1}a_{2}-c_{2}a_{1})^{2} + (a_{1}b_{2}-a_{2}b_{1})^{2}}}$ where $A = \begin{bmatrix} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{bmatrix}$.
Comparing the equations,we get:
$(x_{1}, y_{1}, z_{1}) = (-1, -1, -1)$ and $(a_{1}, b_{1}, c_{1}) = (7, -6, 1)$.
$(x_{2}, y_{2}, z_{2}) = (3, 5, 7)$ and $(a_{2}, b_{2}, c_{2}) = (1, -2, 1)$.
Calculating the determinant:
$\det(A) = \begin{vmatrix} 3-(-1) & 5-(-1) & 7-(-1) \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = \begin{vmatrix} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix}$
$= 4(-6+2) - 6(7-1) + 8(-14+6) = 4(-4) - 6(6) + 8(-8) = -16 - 36 - 64 = -116$.
Calculating the denominator:
$\sqrt{(-6+2)^{2} + (1+7)^{2} + (-14+6)^{2}} = \sqrt{(-4)^{2} + (8)^{2} + (-8)^{2}} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12$.
Wait,re-calculating the cross product vector $\vec{n} = \vec{v_{1}} \times \vec{v_{2}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(-6+2) - \hat{j}(7-1) + \hat{k}(-14+6) = -4\hat{i} - 6\hat{j} - 8\hat{k}$.
Magnitude $|\vec{n}| = \sqrt{(-4)^{2} + (-6)^{2} + (-8)^{2}} = \sqrt{16 + 36 + 64} = \sqrt{116} = 2\sqrt{29}$.
Shortest distance $d = \frac{|-116|}{2\sqrt{29}} = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}$ units.

(A) The given lines are $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$.
The shortest distance $d$ between two skew lines is given by the formula:
$d = \left| \frac{(\vec{b}_{1} \times \vec{b}_{2}) \cdot (\vec{a}_{2} - \vec{a}_{1})}{|\vec{b}_{1} \times \vec{b}_{2}|} \right|$
Comparing the given equations,we have:
$\vec{a}_{1} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b}_{1} = \hat{i} - 3\hat{j} + 2\hat{k}$
$\vec{a}_{2} = 4\hat{i} + 5\hat{j} + 6\hat{k}$,$\vec{b}_{2} = 2\hat{i} + 3\hat{j} + \hat{k}$
Calculate $\vec{a}_{2} - \vec{a}_{1} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}$.
Calculate the cross product $\vec{b}_{1} \times \vec{b}_{2}$:
$\vec{b}_{1} \times \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(-3-6) - \hat{j}(1-4) + \hat{k}(3 - (-6)) = -9\hat{i} + 3\hat{j} + 9\hat{k}$.
Calculate the magnitude $|\vec{b}_{1} \times \vec{b}_{2}| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19}$.
Calculate the dot product $(\vec{b}_{1} \times \vec{b}_{2}) \cdot (\vec{a}_{2} - \vec{a}_{1}) = (-9\hat{i} + 3\hat{j} + 9\hat{k}) \cdot (3\hat{i} + 3\hat{j} + 3\hat{k}) = -27 + 9 + 27 = 9$.
Thus,$d = \left| \frac{9}{3\sqrt{19}} \right| = \frac{3}{\sqrt{19}}$ units.

(A) The given lines are $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$.
Rewriting the equations in the form $\vec{r} = \vec{a} + \lambda \vec{b}$:
Line $1$: $\vec{r} = (\hat{i} - 2\hat{j} + 3\hat{k}) + t(-\hat{i} + \hat{j} - 2\hat{k})$
Here,$\vec{a_1} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b_1} = -\hat{i} + \hat{j} - 2\hat{k}$.
Line $2$: $\vec{r} = (\hat{i} - \hat{j} - \hat{k}) + s(\hat{i} + 2\hat{j} - 2\hat{k})$
Here,$\vec{a_2} = \hat{i} - \hat{j} - \hat{k}$ and $\vec{b_2} = \hat{i} + 2\hat{j} - 2\hat{k}$.
The shortest distance $d$ between two skew lines is given by $d = \left| \frac{(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})}{|\vec{b_1} \times \vec{b_2}|} \right|$.
First,calculate $\vec{b_1} \times \vec{b_2}$:
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} = \hat{i}(-2 + 4) - \hat{j}(2 + 2) + \hat{k}(-2 - 1) = 2\hat{i} - 4\hat{j} - 3\hat{k}$.
Magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$.
Next,calculate $\vec{a_2} - \vec{a_1} = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = 0\hat{i} + \hat{j} - 4\hat{k}$.
Now,$(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) = (2\hat{i} - 4\hat{j} - 3\hat{k}) \cdot (0\hat{i} + \hat{j} - 4\hat{k}) = (2)(0) + (-4)(1) + (-3)(-4) = 0 - 4 + 12 = 8$.
Thus,$d = \left| \frac{8}{\sqrt{29}} \right| = \frac{8}{\sqrt{29}}$ units.

(A) The vector equation of the line passing through points $A(3, 4, 1)$ and $B(5, 1, 6)$ is given by $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.
Substituting the coordinates,we get $\vec{r} = (3\hat{i} + 4\hat{j} + \hat{k}) + \lambda((5-3)\hat{i} + (1-4)\hat{j} + (6-1)\hat{k})$.
This simplifies to $\vec{r} = (3 + 2\lambda)\hat{i} + (4 - 3\lambda)\hat{j} + (1 + 5\lambda)\hat{k}$.
Since the line crosses the $XY$-plane,the $z$-coordinate of any point on the line must be $0$.
Therefore,$1 + 5\lambda = 0$,which gives $\lambda = -\frac{1}{5}$.
Substituting $\lambda = -\frac{1}{5}$ into the expressions for $x$ and $y$:
$x = 3 + 2(-\frac{1}{5}) = 3 - \frac{2}{5} = \frac{13}{5}$.
$y = 4 - 3(-\frac{1}{5}) = 4 + \frac{3}{5} = \frac{23}{5}$.
Thus,the coordinates of the point are $\left( \frac{13}{5}, \frac{23}{5}, 0 \right)$.

(N/A) Let $OA$ be the line joining the origin $O(0,0,0)$ and the point $A(2,1,1).$
The direction ratios of line $OA$ are $(2-0, 1-0, 1-0),$ which are $2, 1, 1.$
Let $BC$ be the line joining the points $B(3,5,-1)$ and $C(4,3,-1).$
The direction ratios of line $BC$ are $(4-3, 3-5, -1-(-1)),$ which are $1, -2, 0.$
Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0.$
Substituting the values: $(2)(1) + (1)(-2) + (1)(0) = 2 - 2 + 0 = 0.$
Since the sum of the products of the direction ratios is $0,$ the line $OA$ is perpendicular to the line $BC.$

(A) line parallel to the $x$-axis has direction ratios proportional to $(1, 0, 0)$.
Since the line passes through the origin $(0, 0, 0)$,its parametric equations are $x = k, y = 0, z = 0$,where $k$ is a parameter.
In symmetric form,this is represented as $\frac{x-0}{1} = \frac{y-0}{0} = \frac{z-0}{0}$,which simplifies to $y=0$ and $z=0$.
Thus,the equation of the line is $y=0, z=0$.

(A) The coordinates of the points are $A(1, 2, 3), B(4, 5, 7), C(-4, 3, -6),$ and $D(2, 9, 2)$.
The direction ratios of line $AB$ are $(4-1, 5-2, 7-3) = (3, 3, 4)$.
The direction ratios of line $CD$ are $(2-(-4), 9-3, 2-(-6)) = (6, 6, 8)$.
Let the direction ratios of $AB$ be $(a_1, b_1, c_1) = (3, 3, 4)$ and $CD$ be $(a_2, b_2, c_2) = (6, 6, 8)$.
We observe that $\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$,$\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$,and $\frac{c_1}{c_2} = \frac{4}{8} = \frac{1}{2}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,the lines $AB$ and $CD$ are parallel.
Therefore,the angle between the parallel lines $AB$ and $CD$ is $0^\circ$.

(A) The direction ratios of the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are $(-3, 2k, 2)$ and $(3k, 1, -5)$ respectively.
Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Substituting the given values,we get:
$-3(3k) + (2k)(1) + (2)(-5) = 0$
$-9k + 2k - 10 = 0$
$-7k - 10 = 0$
$-7k = 10$
$k = -\frac{10}{7}$
Thus,the value of $k$ is $-\frac{10}{7}$.

(A) The given lines are:
$\vec{r} = 6\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})$ ...........$(1)$
$\vec{r} = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})$ ...........$(2)$
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:
$d = \left| \frac{(\vec{b}_{1} \times \vec{b}_{2}) \cdot (\vec{a}_{2} - \vec{a}_{1})}{|\vec{b}_{1} \times \vec{b}_{2}|} \right|$ ...........$(3)$
Comparing the given equations with the standard form,we get:
$\vec{a}_{1} = 6\hat{i} + 2\hat{j} + 2\hat{k}$,$\vec{b}_{1} = \hat{i} - 2\hat{j} + 2\hat{k}$
$\vec{a}_{2} = -4\hat{i} - \hat{k}$,$\vec{b}_{2} = 3\hat{i} - 2\hat{j} - 2\hat{k}$
Now,calculate $\vec{a}_{2} - \vec{a}_{1} = (-4\hat{i} - \hat{k}) - (6\hat{i} + 2\hat{j} + 2\hat{k}) = -10\hat{i} - 2\hat{j} - 3\hat{k}$.
Calculate the cross product $\vec{b}_{1} \times \vec{b}_{2}$:
$\vec{b}_{1} \times \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} = \hat{i}(4 - (-4)) - \hat{j}(-2 - 6) + \hat{k}(-2 - (-6)) = 8\hat{i} + 8\hat{j} + 4\hat{k}$.
The magnitude $|\vec{b}_{1} \times \vec{b}_{2}| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$.
Now,calculate the dot product $(\vec{b}_{1} \times \vec{b}_{2}) \cdot (\vec{a}_{2} - \vec{a}_{1})$:
$(8\hat{i} + 8\hat{j} + 4\hat{k}) \cdot (-10\hat{i} - 2\hat{j} - 3\hat{k}) = (8)(-10) + (8)(-2) + (4)(-3) = -80 - 16 - 12 = -108$.
Finally,the shortest distance $d = \left| \frac{-108}{12} \right| = |-9| = 9$ units.

(A) The equation of the line passing through 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 $(5, 1, 6)$ and $(3, 4, 1)$ into the formula:
$\frac{x-5}{3-5} = \frac{y-1}{4-1} = \frac{z-6}{1-6}$
$\Rightarrow \frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} = k$ (where $k$ is a constant).
Expressing $x, y, z$ in terms of $k$:
$x = 5 - 2k$
$y = 3k + 1$
$z = 6 - 5k$
Any point on this line is of the form $(5 - 2k, 3k + 1, 6 - 5k)$.
The equation of the $YZ$-plane is $x = 0$.
Since the line crosses the $YZ$-plane,we set the $x$-coordinate to $0$:
$5 - 2k = 0 \Rightarrow k = \frac{5}{2}$.
Now,substitute $k = \frac{5}{2}$ into the expressions for $y$ and $z$:
$y = 3\left(\frac{5}{2}\right) + 1 = \frac{15}{2} + 1 = \frac{17}{2}$.
$z = 6 - 5\left(\frac{5}{2}\right) = 6 - \frac{25}{2} = \frac{12 - 25}{2} = -\frac{13}{2}$.
Thus,the coordinates of the required point are $\left(0, \frac{17}{2}, -\frac{13}{2}\right)$.

(A) The equation of the line passing through points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.
The line passing through $(5, 1, 6)$ and $(3, 4, 1)$ is given by $\frac{x-5}{3-5} = \frac{y-1}{4-1} = \frac{z-6}{1-6}$.
This simplifies to $\frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} = k$.
Any point on this line is of the form $(5-2k, 3k+1, 6-5k)$.
Since the line crosses the $ZX$-plane,the $y$-coordinate must be $0$.
Setting $3k+1 = 0$,we get $k = -\frac{1}{3}$.
Substituting $k = -\frac{1}{3}$ into the coordinates:
$x = 5 - 2(-\frac{1}{3}) = 5 + \frac{2}{3} = \frac{17}{3}$.
$y = 3(-\frac{1}{3}) + 1 = 0$.
$z = 6 - 5(-\frac{1}{3}) = 6 + \frac{5}{3} = \frac{23}{3}$.
Thus,the required point is $\left(\frac{17}{3}, 0, \frac{23}{3}\right)$.

(A) Let the required line be parallel to the vector $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.
The position vector of the point $(1, 2, -4)$ is $\vec{a} = \hat{i} + 2\hat{j} - 4\hat{k}$.
The equation of the line passing through $\vec{a}$ and parallel to $\vec{b}$ is $\vec{r} = \vec{a} + \lambda\vec{b}$.
The given lines are parallel to vectors $\vec{v}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k}$ and $\vec{v}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k}$.
Since the required line is perpendicular to both lines,its direction vector $\vec{b}$ is given by the cross product $\vec{v}_1 \times \vec{v}_2$.
$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) = 24\hat{i} + 36\hat{j} + 72\hat{k}$.
Dividing by $12$,we get the direction ratios as $(2, 3, 6)$.
Thus,$\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.
The vector equation is $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$.

(N/A) Let the given point be $P(-2, 4, -5)$.
Any point $Q$ on the line $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} = \lambda$ is given by $Q(3\lambda - 3, 5\lambda + 4, 6\lambda - 8)$.
The vector $\overrightarrow{PQ}$ is given by $\overrightarrow{PQ} = (3\lambda - 3 - (-2))\hat{i} + (5\lambda + 4 - 4)\hat{j} + (6\lambda - 8 - (-5))\hat{k} = (3\lambda - 1)\hat{i} + 5\lambda\hat{j} + (6\lambda - 3)\hat{k}$.
Since $\overrightarrow{PQ}$ is perpendicular to the line with direction vector $\vec{v} = 3\hat{i} + 5\hat{j} + 6\hat{k}$,their dot product must be zero:
$\overrightarrow{PQ} \cdot \vec{v} = 0$
$3(3\lambda - 1) + 5(5\lambda) + 6(6\lambda - 3) = 0$
$9\lambda - 3 + 25\lambda + 36\lambda - 18 = 0$
$70\lambda - 21 = 0 \implies \lambda = \frac{21}{70} = \frac{3}{10}$.
Substituting $\lambda = \frac{3}{10}$ into $\overrightarrow{PQ}$:
$\overrightarrow{PQ} = (3(\frac{3}{10}) - 1)\hat{i} + 5(\frac{3}{10})\hat{j} + (6(\frac{3}{10}) - 3)\hat{k} = (\frac{9-10}{10})\hat{i} + \frac{15}{10}\hat{j} + (\frac{18-30}{10})\hat{k} = -\frac{1}{10}\hat{i} + \frac{15}{10}\hat{j} - \frac{12}{10}\hat{k}$.
The distance is the magnitude of $\overrightarrow{PQ}$:
$|\overrightarrow{PQ}| = \sqrt{(-\frac{1}{10})^2 + (\frac{15}{10})^2 + (-\frac{12}{10})^2} = \sqrt{\frac{1 + 225 + 144}{100}} = \sqrt{\frac{370}{100}} = \sqrt{\frac{37}{10}} \text{ units.}$

(A) Given equations are $3l+m+5n=0$ $(1)$ and $6mn-2nl+5lm=0$ $(2)$.
From $(1)$,$m = -(3l+5n)$.
Substituting this into $(2)$: $6n(-(3l+5n)) - 2nl + 5l(-(3l+5n)) = 0$.
$-18ln - 30n^2 - 2nl - 15l^2 - 25ln = 0$.
$-15l^2 - 45ln - 30n^2 = 0$.
Dividing by $-15$: $l^2 + 3ln + 2n^2 = 0$.
$(l+n)(l+2n) = 0$.
Case $1$: $l = -n$. Then $m = -(3(-n)+5n) = -2n$. Direction ratios are $(-n, -2n, n)$,i.e.,$(1, 2, -1)$.
Case $2$: $l = -2n$. Then $m = -(3(-2n)+5n) = n$. Direction ratios are $(-2n, n, n)$,i.e.,$(-2, 1, 1)$.
Let $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = -2\hat{i} + \hat{j} + \hat{k}$.
$\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|} = \frac{|(1)(-2) + (2)(1) + (-1)(1)|}{\sqrt{1^2+2^2+(-1)^2} \sqrt{(-2)^2+1^2+1^2}} = \frac{|-2+2-1|}{\sqrt{6}\sqrt{6}} = \frac{1}{6}$.
Thus,$\theta = \cos^{-1}\left(\frac{1}{6}\right)$.

(N/A) Let $L$ be the foot of the perpendicular drawn from point $A(1, 8, 4)$ to the line passing through $B(0, -1, 3)$ and $C(2, -3, -1)$.
The direction ratios of the line $BC$ are $(2-0, -3-(-1), -1-3)$,which are $(2, -2, -4)$.
The equation of the line $BC$ passing through $B(0, -1, 3)$ with direction ratios $(2, -2, -4)$ is given by:
$\frac{x-0}{2} = \frac{y+1}{-2} = \frac{z-3}{-4} = \lambda$
Any point $L$ on the line $BC$ can be represented as:
$L = (2\lambda, -2\lambda-1, -4\lambda+3)$
The direction ratios of the line $AL$ are:
$(2\lambda-1, -2\lambda-1-8, -4\lambda+3-4) = (2\lambda-1, -2\lambda-9, -4\lambda-1)$
Since $AL \perp BC$,the dot product of their direction ratios is zero:
$2(2\lambda-1) - 2(-2\lambda-9) - 4(-4\lambda-1) = 0$
$4\lambda - 2 + 4\lambda + 18 + 16\lambda + 4 = 0$
$24\lambda + 20 = 0$
$24\lambda = -20$
$\lambda = -\frac{20}{24} = -\frac{5}{6}$
Substituting $\lambda = -\frac{5}{6}$ into the coordinates of $L$:
$x = 2(-\frac{5}{6}) = -\frac{5}{3}$
$y = -2(-\frac{5}{6}) - 1 = \frac{5}{3} - 1 = \frac{2}{3}$
$z = -4(-\frac{5}{6}) + 3 = \frac{10}{3} + 3 = \frac{19}{3}$
Thus,the coordinates of the foot of the perpendicular are $(-\frac{5}{3}, \frac{2}{3}, \frac{19}{3})$.

(D) Let $P(1, 6, 3)$ be the given point and let $L$ be the foot of the perpendicular from $P$ to the given line.
The coordinates of a general point on the given line are $\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-2}{3} = \lambda$,i.e.,$x = \lambda, y = 2\lambda + 1, z = 3\lambda + 2$.
If the coordinates of $L$ are $(\lambda, 2\lambda + 1, 3\lambda + 2)$,then the direction ratios of $PL$ are $(\lambda - 1, 2\lambda + 1 - 6, 3\lambda + 2 - 3)$,which simplifies to $(\lambda - 1, 2\lambda - 5, 3\lambda - 1)$.
The direction ratios of the given line are $(1, 2, 3)$. Since $PL$ is perpendicular to the line,the dot product of their direction ratios must be zero:
$1(\lambda - 1) + 2(2\lambda - 5) + 3(3\lambda - 1) = 0$
$\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0$
$14\lambda - 14 = 0 \Rightarrow \lambda = 1$.
Substituting $\lambda = 1$ into the coordinates of $L$,we get $L(1, 2(1) + 1, 3(1) + 2) = (1, 3, 5)$.
Let $Q(x_1, y_1, z_1)$ be the image of $P(1, 6, 3)$ in the given line. Then $L$ is the midpoint of $PQ$. Therefore:
$\frac{x_1 + 1}{2} = 1 \Rightarrow x_1 = 1$
$\frac{y_1 + 6}{2} = 3 \Rightarrow y_1 = 0$
$\frac{z_1 + 3}{2} = 5 \Rightarrow z_1 = 7$
Hence,the image of $(1, 6, 3)$ in the given line is $(1, 0, 7)$.

(A) Let the equation of the given line be $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=\lambda$.
Any point on this line is $P(2\lambda+3, \lambda+3, \lambda)$.
The direction ratios of the line are $(2, 1, 1)$.
Since the required lines pass through the origin $(0, 0, 0)$,the direction ratios of the required lines are $(2\lambda+3, \lambda+3, \lambda)$.
The angle $\theta$ between the two lines is given by $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Given $\theta = \frac{\pi}{3}$,so $\cos \frac{\pi}{3} = \frac{1}{2}$.
Substituting the values:
$\frac{1}{2} = \frac{|2(2\lambda+3) + 1(\lambda+3) + 1(\lambda)|}{\sqrt{2^2+1^2+1^2} \sqrt{(2\lambda+3)^2 + (\lambda+3)^2 + \lambda^2}}$
$\frac{1}{2} = \frac{|4\lambda+6+\lambda+3+\lambda|}{\sqrt{6} \sqrt{4\lambda^2+12\lambda+9 + \lambda^2+6\lambda+9 + \lambda^2}}$
$\frac{1}{2} = \frac{|6\lambda+9|}{\sqrt{6} \sqrt{6\lambda^2+18\lambda+18}}$
Squaring both sides:
$\frac{1}{4} = \frac{(6\lambda+9)^2}{6(6\lambda^2+18\lambda+18)}$
$6(6\lambda^2+18\lambda+18) = 4(36\lambda^2+108\lambda+81)$
$36\lambda^2+108\lambda+108 = 144\lambda^2+432\lambda+324$
$108\lambda^2+324\lambda+216 = 0$
Dividing by $108$:
$\lambda^2+3\lambda+2 = 0 \Rightarrow (\lambda+1)(\lambda+2) = 0$.
So,$\lambda = -1$ or $\lambda = -2$.
For $\lambda = -1$,the direction ratios are $(2(-1)+3, -1+3, -1) = (1, 2, -1)$.
For $\lambda = -2$,the direction ratios are $(2(-2)+3, -2+3, -2) = (-1, 1, -2)$.
Thus,the equations of the lines are $\frac{x}{1} = \frac{y}{2} = \frac{z}{-1}$ and $\frac{x}{-1} = \frac{y}{1} = \frac{z}{-2}$.

(N/A) We have the equation of the line as $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$.
Rewriting it in standard form: $\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}=\lambda$.
Thus,any point on the line is given by $x = -2\lambda + 4$,$y = 6\lambda$,and $z = -3\lambda + 1$.
Let $L$ be the foot of the perpendicular from $P(2, 3, -8)$ to the line. So,the coordinates of $L$ are $(4-2\lambda, 6\lambda, 1-3\lambda)$.
The direction ratios of the line $PL$ are $(4-2\lambda-2, 6\lambda-3, 1-3\lambda+8)$,which simplifies to $(2-2\lambda, 6\lambda-3, 9-3\lambda)$.
The direction ratios of the given line are $(-2, 6, -3)$.
Since $PL$ is perpendicular to the line,the dot product of their direction ratios is zero:
$-2(2-2\lambda) + 6(6\lambda-3) - 3(9-3\lambda) = 0$.
$-4 + 4\lambda + 36\lambda - 18 - 27 + 9\lambda = 0$.
$49\lambda - 49 = 0 \Rightarrow \lambda = 1$.
Substituting $\lambda = 1$ into the coordinates of $L$,we get $L = (4-2(1), 6(1), 1-3(1)) = (2, 6, -2)$.
The perpendicular distance $PL$ is the distance between $P(2, 3, -8)$ and $L(2, 6, -2)$:
$PL = \sqrt{(2-2)^2 + (6-3)^2 + (-2 - (-8))^2} = \sqrt{0^2 + 3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$ units.

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,the position vector of the point $(1, -2, 3)$ is $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}$.
The line is parallel to the vector $\vec{b} = 3 \hat{i} - 2 \hat{j} + 6 \hat{k}$.
Substituting these values into the formula,we get:
$\vec{r} = (\hat{i} - 2 \hat{j} + 3 \hat{k}) + \lambda (3 \hat{i} - 2 \hat{j} + 6 \hat{k})$,where $\lambda$ is a scalar.

(A) Let the equations of the lines be:
$L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} = \lambda$
$L_2: \frac{x-4}{5}=\frac{y-1}{2}=\frac{z-0}{1} = \mu$
Any point on $L_1$ is $(2\lambda+1, 3\lambda+2, 4\lambda+3)$ and any point on $L_2$ is $(5\mu+4, 2\mu+1, \mu)$.
If the lines intersect,there exist $\lambda$ and $\mu$ such that:
$2\lambda+1 = 5\mu+4 \Rightarrow 2\lambda - 5\mu = 3$ $(1)$
$3\lambda+2 = 2\mu+1 \Rightarrow 3\lambda - 2\mu = -1$ $(2)$
$4\lambda+3 = \mu$ $(3)$
Substitute $(3)$ into $(1)$:
$2\lambda - 5(4\lambda+3) = 3$
$2\lambda - 20\lambda - 15 = 3$
$-18\lambda = 18 \Rightarrow \lambda = -1$
Using $\lambda = -1$ in $(3)$:
$\mu = 4(-1)+3 = -1$
Check these values in $(2)$:
$3(-1) - 2(-1) = -3 + 2 = -1$. This satisfies equation $(2)$.
Since the values satisfy all equations,the lines intersect.
Point of intersection using $\lambda = -1$:
$x = 2(-1)+1 = -1$
$y = 3(-1)+2 = -1$
$z = 4(-1)+3 = -1$
Thus,the point of intersection is $(-1, -1, -1)$.

(N/A) The given lines are $\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1}$ and $\vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}$.
Here,$\vec{b}_{1}=2 \hat{i}+\hat{j}+2 \hat{k}$ and $\vec{b}_{2}=6 \hat{i}+3 \hat{j}+2 \hat{k}$.
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: $\vec{b}_{1} \cdot \vec{b}_{2} = (2)(6) + (1)(3) + (2)(2) = 12 + 3 + 4 = 19$.
Calculating the magnitudes: $|\vec{b}_{1}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.
$|\vec{b}_{2}| = \sqrt{6^2 + 3^2 + 2^2} = \sqrt{36+9+4} = \sqrt{49} = 7$.
Thus,$\cos \theta = \frac{19}{3 \times 7} = \frac{19}{21}$.
Therefore,$\theta = \cos^{-1} \left( \frac{19}{21} \right)$.

(A) The Cartesian equation of a line passing through two points $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ is given by $\frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}$.
The equation of the line passing through $A(0,-1,-1)$ and $B(4,5,1)$ is $\frac{x-0}{4-0} = \frac{y+1}{5+1} = \frac{z+1}{1+1}$,which simplifies to $\frac{x}{4} = \frac{y+1}{6} = \frac{z+1}{2} = \lambda$ (say). Any point on this line is $(4\lambda, 6\lambda-1, 2\lambda-1)$.
The equation of the line passing through $C(3,9,4)$ and $D(-4,4,4)$ is $\frac{x-3}{-4-3} = \frac{y-9}{4-9} = \frac{z-4}{4-4}$,which simplifies to $\frac{x-3}{-7} = \frac{y-9}{-5} = \frac{z-4}{0} = \mu$ (say). Any point on this line is $(-7\mu+3, -5\mu+9, 4)$.
For the lines to intersect,there must exist $\lambda$ and $\mu$ such that the coordinates are equal:
$2\lambda-1 = 4 \Rightarrow 2\lambda = 5 \Rightarrow \lambda = 2.5$.
Substituting $\lambda = 2.5$ into the first two coordinates: $x = 4(2.5) = 10$,$y = 6(2.5)-1 = 14$.
Now,for the second line: $-7\mu+3 = 10 \Rightarrow -7\mu = 7 \Rightarrow \mu = -1$.
Checking $y$: $-5(-1)+9 = 5+9 = 14$. Since the coordinates match,the lines intersect.

(A) The given equations of the first line are $x=p y+q$ and $z=r y+s$.
This can be rewritten as $\frac{x-q}{p} = y = \frac{z-s}{r}$.
Thus,the direction ratios of the first line are $(p, 1, r)$.
Similarly,the equations of the second line are $x=p^{\prime} y+q^{\prime}$ and $z=r^{\prime} y+s^{\prime}$.
This can be rewritten as $\frac{x-q^{\prime}}{p^{\prime}} = y = \frac{z-s^{\prime}}{r^{\prime}}$.
Thus,the direction ratios of the second line are $(p^{\prime}, 1, r^{\prime})$.
Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Substituting the direction ratios,we get $(p)(p^{\prime}) + (1)(1) + (r)(r^{\prime}) = 0$.
This simplifies to $p p^{\prime} + r r^{\prime} + 1 = 0$.
Hence,the lines are perpendicular if $p p^{\prime} + r r^{\prime} + 1 = 0$.

(C) Let the equation of the line be $\frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9} = \lambda$.
Then,any point $L$ on the line is given by $x = \lambda - 5$,$y = 4\lambda - 3$,$z = 6 - 9\lambda$.
Let $P$ be the point $(2, 4, -1)$. The direction ratios of the line segment $PL$ are $(\lambda - 5 - 2, 4\lambda - 3 - 4, 6 - 9\lambda - (-1))$,which simplifies to $(\lambda - 7, 4\lambda - 7, 7 - 9\lambda)$.
Since $PL$ is perpendicular to the given line,the dot product of the direction ratios of $PL$ and the line $(1, 4, -9)$ must be zero:
$1(\lambda - 7) + 4(4\lambda - 7) - 9(7 - 9\lambda) = 0$.
Expanding this,we get $\lambda - 7 + 16\lambda - 28 - 63 + 81\lambda = 0$.
Combining like terms,$98\lambda - 98 = 0$,which gives $\lambda = 1$.
Substituting $\lambda = 1$ back into the coordinates of $L$,we get $L = (1 - 5, 4(1) - 3, 6 - 9(1)) = (-4, 1, -3)$.
The distance $PL$ is $\sqrt{(-4 - 2)^2 + (1 - 4)^2 + (-3 - (-1))^2} = \sqrt{(-6)^2 + (-3)^2 + (-2)^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$ units.

(C) The given lines are $\vec{r}=(8+3 \lambda) \hat{i}+(-9-16 \lambda) \hat{j}+(10+7 \lambda) \hat{k}$ and $\vec{r}=15 \hat{i}+29 \hat{j}+5 \hat{k}+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k})$.
First line: $\vec{r}=(8 \hat{i}-9 \hat{j}+10 \hat{k})+\lambda(3 \hat{i}-16 \hat{j}+7 \hat{k})$.
So,$\vec{a}_{1}=8 \hat{i}-9 \hat{j}+10 \hat{k}$ and $\vec{b}_{1}=3 \hat{i}-16 \hat{j}+7 \hat{k}$.
Second line: $\vec{a}_{2}=15 \hat{i}+29 \hat{j}+5 \hat{k}$ and $\vec{b}_{2}=3 \hat{i}+8 \hat{j}-5 \hat{k}$.
The shortest distance $d$ is given by $d = \left|\frac{(\vec{b}_{1} \times \vec{b}_{2}) \cdot (\vec{a}_{2}-\vec{a}_{1})}{|\vec{b}_{1} \times \vec{b}_{2}|}\right|$.
Calculate $\vec{b}_{1} \times \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = \hat{i}(80-56) - \hat{j}(-15-21) + \hat{k}(24+48) = 24 \hat{i}+36 \hat{j}+72 \hat{k}$.
Magnitude $|\vec{b}_{1} \times \vec{b}_{2}| = \sqrt{24^2 + 36^2 + 72^2} = \sqrt{576 + 1296 + 5184} = \sqrt{7056} = 84$.
Calculate $\vec{a}_{2}-\vec{a}_{1} = (15-8)\hat{i} + (29-(-9))\hat{j} + (5-10)\hat{k} = 7\hat{i} + 38\hat{j} - 5\hat{k}$.
Dot product $(\vec{b}_{1} \times \vec{b}_{2}) \cdot (\vec{a}_{2}-\vec{a}_{1}) = (24)(7) + (36)(38) + (72)(-5) = 168 + 1368 - 360 = 1176$.
Shortest distance $d = \left|\frac{1176}{84}\right| = 14 \text{ units}$.

(A) We have,$2l + 2m - n = 0 \dots (i)$
And $mn + nl + lm = 0 \dots (ii)$
From $(i)$,$n = 2l + 2m$. Substituting this into $(ii)$:
$m(2l + 2m) + (2l + 2m)l + lm = 0$
$2lm + 2m^2 + 2l^2 + 2lm + lm = 0$
$2l^2 + 5lm + 2m^2 = 0$
$2l^2 + 4lm + lm + 2m^2 = 0$
$2l(l + 2m) + m(l + 2m) = 0$
$(2l + m)(l + 2m) = 0$
Case $1$: $m = -2l$. Then $n = 2l + 2(-2l) = -2l$. The direction ratios are $(l, -2l, -2l)$,which are proportional to $(1, -2, -2)$.
Case $2$: $l = -2m$. Then $n = 2(-2m) + 2m = -2m$. The direction ratios are $(-2m, m, -2m)$,which are proportional to $(2, -1, 2)$.
Let the direction ratios be $a_1, b_1, c_1 = (1, -2, -2)$ and $a_2, b_2, c_2 = (2, -1, 2)$.
The condition for perpendicularity is $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
$(1)(2) + (-2)(-1) + (-2)(2) = 2 + 2 - 4 = 0$.
Since the sum is $0$,the lines are at right angles.

(C) The given lines are $\overrightarrow{r} = \hat{i}(1 + 2\ell) + \hat{j}(-1) + \hat{k}(\ell)$ and $\overrightarrow{r} = \hat{i}(2 + m) + \hat{j}(m - 1) + \hat{k}(-m)$.
For the lines to intersect,there must exist values of $\ell$ and $m$ such that the coordinates are equal:
$1 + 2\ell = 2 + m$ $(i)$
$-1 = m - 1$ $(ii)$
$\ell = -m$ $(iii)$
From equation $(ii)$,we get $m = 0$.
Substituting $m = 0$ into equation $(iii)$,we get $\ell = 0$.
Now,check if these values satisfy equation $(i)$:
$1 + 2(0) = 2 + 0 \implies 1 = 2$,which is a contradiction.
Since the values of $\ell$ and $m$ do not satisfy all three equations simultaneously,the lines do not intersect for any values of $\ell$ and $m$.

(B) Let the given line be $\frac{x+1}{2} = \frac{y-3}{-2} = \frac{z}{-1} = \lambda$.
Any point $R$ on the line is given by $(2\lambda-1, -2\lambda+3, -\lambda)$.
Let $P = (1, 2, -3)$ and $Q = (a, b, c)$ be the image of $P$ in the line.
The vector $\vec{PQ} = (a-1, b-2, c+3)$ must be perpendicular to the line,whose direction ratios are $(2, -2, -1)$.
Thus,$2(a-1) - 2(b-2) - 1(c+3) = 0 \implies 2a - 2b - c = 1$.
Also,the midpoint $R$ of $PQ$ lies on the line:
$R = \left(\frac{a+1}{2}, \frac{b+2}{2}, \frac{c-3}{2}\right)$.
Since $R$ lies on the line,we have:
$\frac{\frac{a+1}{2} + 1}{2} = \frac{\frac{b+2}{2} - 3}{-2} = \frac{\frac{c-3}{2}}{-1} = \lambda$.
Solving for $a, b, c$ in terms of $\lambda$:
$a = 4\lambda - 3, b = -4\lambda + 4, c = -2\lambda + 3$.
Substituting these into the perpendicularity condition $2a - 2b - c = 1$:
$2(4\lambda - 3) - 2(-4\lambda + 4) - (-2\lambda + 3) = 1
\implies 8\lambda - 6 + 8\lambda - 8 + 2\lambda - 3 = 1
\implies 18\lambda = 18 \implies \lambda = 1$.
Thus,$a = 4(1) - 3 = 1$,$b = -4(1) + 4 = 0$,$c = -2(1) + 3 = 1$.
Therefore,$a+b+c = 1 + 0 + 1 = 2$.

(D) The first line is $L_1: \frac{x-1}{0} = \frac{y+1}{-1} = \frac{z}{1}$. $A$ point on $L_1$ is $A(1, -1, 0)$ and its direction vector is $\vec{c} = (0, -1, 1)$.
The second line $L_2$ is the intersection of planes $x+y+z+1=0$ and $2x-y+z+3=0$.
Adding the two equations: $(x+y+z+1) + (2x-y+z+3) = 3x + 2z + 4 = 0$,so $x = \frac{-2z-4}{3}$.
Substituting $x$ into the first plane: $\frac{-2z-4}{3} + y + z + 1 = 0 \Rightarrow y = -z - 1 + \frac{2z+4}{3} = \frac{-3z-3+2z+4}{3} = \frac{-z+1}{3}$.
Thus,$x = \frac{-2z-4}{3}, y = \frac{-z+1}{3}, z = z$. Rewriting in symmetric form: $\frac{x+4/3}{-2/3} = \frac{y-1/3}{-1/3} = \frac{z}{1}$.
$A$ point on $L_2$ is $B(-\frac{4}{3}, \frac{1}{3}, 0)$ and its direction vector is $\vec{d} = (-\frac{2}{3}, -\frac{1}{3}, 1)$.
Vector $\vec{AB} = B - A = (-\frac{4}{3}-1, \frac{1}{3}-(-1), 0-0) = (-\frac{7}{3}, \frac{4}{3}, 0)$.
Cross product $\vec{c} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -1 & 1 \\ -2/3 & -1/3 & 1 \end{vmatrix} = \hat{i}(-1 + 1/3) - \hat{j}(0 + 2/3) + \hat{k}(0 - 2/3) = (-\frac{2}{3}, -\frac{2}{3}, -\frac{2}{3})$.
Magnitude $|\vec{c} \times \vec{d}| = \sqrt{(-\frac{2}{3})^2 + (-\frac{2}{3})^2 + (-\frac{2}{3})^2} = \sqrt{\frac{4}{9} + \frac{4}{9} + \frac{4}{9}} = \sqrt{\frac{12}{9}} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$.
Shortest distance $d = \frac{|\vec{AB} \cdot (\vec{c} \times \vec{d})|}{|\vec{c} \times \vec{d}|} = \frac{|(-\frac{7}{3})(-\frac{2}{3}) + (\frac{4}{3})(-\frac{2}{3}) + (0)(-\frac{2}{3})|}{2/\sqrt{3}} = \frac{|\frac{14}{9} - \frac{8}{9}|}{2/\sqrt{3}} = \frac{6/9}{2/\sqrt{3}} = \frac{2/3}{2/\sqrt{3}} = \frac{1}{\sqrt{3}}$.

(B) Since $(3,5,7)$ lies on $L_{1}$,we have $\frac{3-a}{l}=\frac{5-2}{3}=\frac{7-b}{4}=1$.
From this,$3-a=l \Rightarrow a+l=3$ and $7-b=4 \Rightarrow b=3$.
The vector from $(3,5,7)$ to $(4,3,8)$ is $\vec{v}_{1} = (4-3, 3-5, 8-7) = (1, -2, 1)$.
The direction vector of $L_{1}$ is $\vec{v}_{2} = (l, 3, 4)$.
Since the line segment is perpendicular to $L_{1}$,$\vec{v}_{1} \cdot \vec{v}_{2} = 0 \Rightarrow l - 6 + 4 = 0 \Rightarrow l = 2$.
Thus,$a = 3 - 2 = 1$.
The lines are $L_{1}: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $L_{2}: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$.
Let $A = (1,2,3)$ and $B = (2,4,5)$,so $\vec{AB} = (1,2,2)$.
The direction vectors are $\vec{p} = (2,3,4)$ and $\vec{q} = (3,4,5)$.
$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$.
The shortest distance is $d = \left| \frac{\vec{AB} \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right| = \left| \frac{(1,2,2) \cdot (-1,2,-1)}{\sqrt{(-1)^2 + 2^2 + (-1)^2}} \right| = \left| \frac{-1+4-2}{\sqrt{6}} \right| = \frac{1}{\sqrt{6}}$.

(D) Let the given line be $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2} = r$.
Any point $P$ on $L_1$ is given by $P(2r+1, 3r-1, -2r+1)$.
The direction vector of $L_1$ is $\vec{v_1} = 2\hat{i} + 3\hat{j} - 2\hat{k}$.
The vector $\vec{QP}$ is $(2r+1-0)\hat{i} + (3r-1-1)\hat{j} + (-2r+1-2)\hat{k} = (2r+1)\hat{i} + (3r-2)\hat{j} + (-2r-1)\hat{k}$.
Since the required line is perpendicular to $L_1$,$\vec{QP} \cdot \vec{v_1} = 0$.
$2(2r+1) + 3(3r-2) - 2(-2r-1) = 0$.
$4r + 2 + 9r - 6 + 4r + 2 = 0$.
$17r - 2 = 0 \Rightarrow r = \frac{2}{17}$.
The direction ratios of the line $QP$ are proportional to the components of $\vec{QP}$ at $r = \frac{2}{17}$.
$vec{QP} = (2(\frac{2}{17})+1)\hat{i} + (3(\frac{2}{17})-2)\hat{j} + (-2(\frac{2}{17})-1)\hat{k} = \frac{21}{17}\hat{i} - \frac{28}{17}\hat{j} - \frac{21}{17}\hat{k}$.
Dividing by $\frac{7}{17}$,the direction ratios are $(3, -4, -3)$.
Thus,the equation of the line passing through $(0,1,2)$ with direction ratios $(3, -4, -3)$ is $\frac{x-0}{3} = \frac{y-1}{-4} = \frac{z-2}{-3}$,which is equivalent to $\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}$.

(D) The given lines are $L_1: \frac{x-\lambda}{1} = \frac{y-1/2}{1/2} = \frac{z-0}{-1/2}$ and $L_2: \frac{x-0}{1} = \frac{y+2\lambda}{1} = \frac{z-\lambda}{1}$.
The points on the lines are $a_1 = (\lambda, 1/2, 0)$ and $a_2 = (0, -2\lambda, \lambda)$.
The direction vectors are $b_1 = (1, 1/2, -1/2)$ and $b_2 = (1, 1, 1)$.
The cross product $b_1 \times b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1/2 & -1/2 \\ 1 & 1 & 1 \end{vmatrix} = \hat{i}(1/2 + 1/2) - \hat{j}(1 + 1/2) + \hat{k}(1 - 1/2) = \hat{i} - \frac{3}{2}\hat{j} + \frac{1}{2}\hat{k} = \frac{1}{2}(2\hat{i} - 3\hat{j} + \hat{k})$.
The magnitude $|b_1 \times b_2| = \frac{1}{2}\sqrt{2^2 + (-3)^2 + 1^2} = \frac{\sqrt{14}}{2}$.
The vector $a_2 - a_1 = (0-\lambda, -2\lambda-1/2, \lambda-0) = (-\lambda, -2\lambda-1/2, \lambda)$.
The shortest distance $d = \frac{|(a_2 - a_1) \cdot (b_1 \times b_2)|}{|b_1 \times b_2|} = \frac{|-\lambda(1) - (2\lambda+1/2)(-3/2) + \lambda(1/2)|}{\sqrt{14}/2} = \frac{|-\lambda + 3\lambda + 3/4 + \lambda/2|}{\sqrt{14}/2} = \frac{|5\lambda/2 + 3/4|}{\sqrt{14}/2} = \frac{|10\lambda + 3|}{2\sqrt{14}}$.
Given $d = \frac{\sqrt{7}}{2\sqrt{2}} = \frac{\sqrt{14}}{4}$.
So,$\frac{|10\lambda + 3|}{2\sqrt{14}} = \frac{\sqrt{14}}{4} \implies |10\lambda + 3| = \frac{14}{2} = 7$.
Case $1$: $10\lambda + 3 = 7 \implies 10\lambda = 4 \implies \lambda = 0.4$ (Not an integer).
Case $2$: $10\lambda + 3 = -7 \implies 10\lambda = -10 \implies \lambda = -1$.
Thus,$|\lambda| = |-1| = 1$.

(C) The line is given by the intersection of two planes: $3y - 2z - 1 = 0$ and $3x - z + 4 = 0$.
The direction vector $\vec{v}$ of the line is the cross product of the normals to the planes,$\vec{n_1} = (0, 3, -2)$ and $\vec{n_2} = (3, 0, -1)$.
$\vec{v} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & -2 \\ 3 & 0 & -1 \end{vmatrix} = \hat{i}(-3) - \hat{j}(6) + \hat{k}(-9) = (-3, -6, -9)$.
We can simplify the direction ratios to $(1, 2, 3)$.
To find a point on the line,let $z = k$. Then $3y = 2k + 1 \Rightarrow y = \frac{2k+1}{3}$ and $3x = k - 4 \Rightarrow x = \frac{k-4}{3}$.
Setting $k = 1$,we get the point $P = (-1, 1, 1)$.
The line equation is $\frac{x+1}{1} = \frac{y-1}{2} = \frac{z-1}{3} = \lambda$.
Any point $Q$ on the line is $(\lambda - 1, 2\lambda + 1, 3\lambda + 1)$.
The vector $\vec{PQ} = (\lambda - 1 - 2, 2\lambda + 1 - (-1), 3\lambda + 1 - 6) = (\lambda - 3, 2\lambda + 2, 3\lambda - 5)$.
Since $\vec{PQ}$ is perpendicular to the line direction $(1, 2, 3)$,their dot product is zero:
$1(\lambda - 3) + 2(2\lambda + 2) + 3(3\lambda - 5) = 0$.
$\lambda - 3 + 4\lambda + 4 + 9\lambda - 15 = 0 \Rightarrow 14\lambda - 14 = 0 \Rightarrow \lambda = 1$.
The point $Q$ is $(1-1, 2+1, 3+1) = (0, 3, 4)$.
The distance from $(2, -1, 6)$ to $(0, 3, 4)$ is $\sqrt{(0-2)^2 + (3 - (-1))^2 + (4-6)^2} = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6}$.

(D) 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,$\vec{a}_1 = \alpha \hat{i} + 2 \hat{j} + 2 \hat{k}$,$\vec{b}_1 = \hat{i} - 2 \hat{j} + 2 \hat{k}$,$\vec{a}_2 = -4 \hat{i} - \hat{k}$,and $\vec{b}_2 = 3 \hat{i} - 2 \hat{j} - 2 \hat{k}$.
First,calculate $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} = \hat{i}(4 - (-4)) - \hat{j}(-2 - 6) + \hat{k}(-2 - (-6)) = 8 \hat{i} + 8 \hat{j} + 4 \hat{k}$.
The magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$.
Now,$\vec{a}_2 - \vec{a}_1 = (-4 - \alpha) \hat{i} - 2 \hat{j} - 3 \hat{k}$.
The shortest distance is $9 = \frac{|((-4 - \alpha) \hat{i} - 2 \hat{j} - 3 \hat{k}) \cdot (8 \hat{i} + 8 \hat{j} + 4 \hat{k})|}{12}$.
$9 = \frac{|8(-4 - \alpha) - 16 - 12|}{12} \implies 108 = |-32 - 8\alpha - 28| = |-60 - 8\alpha|$.
Since $\alpha > 0$,$|-60 - 8\alpha| = 60 + 8\alpha$.
$60 + 8\alpha = 108 \implies 8\alpha = 48 \implies \alpha = 6$.

(C) The given lines are:
$L_{1}: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$. Point $P_{1} = (1, 2, 1)$,direction vector $\vec{v}_{1} = 2\hat{i} + \hat{j} + 3\hat{k}$.
$L_{2}: \frac{x-2}{1} = \frac{y-\lambda}{2} = \frac{z-3}{4}$. Point $P_{2} = (2, \lambda, 3)$,direction vector $\vec{v}_{2} = \hat{i} + 2\hat{j} + 4\hat{k}$.
Let $\vec{a} = P_{2} - P_{1} = (2-1)\hat{i} + (\lambda-2)\hat{j} + (3-1)\hat{k} = \hat{i} + (\lambda-2)\hat{j} + 2\hat{k}$.
The shortest distance $d$ is given by $d = \frac{|\vec{a} \cdot (\vec{v}_{1} \times \vec{v}_{2})|}{|\vec{v}_{1} \times \vec{v}_{2}|}$.
First,calculate $\vec{v}_{1} \times \vec{v}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{vmatrix} = \hat{i}(4-6) - \hat{j}(8-3) + \hat{k}(4-1) = -2\hat{i} - 5\hat{j} + 3\hat{k}$.
Magnitude $|\vec{v}_{1} \times \vec{v}_{2}| = \sqrt{(-2)^2 + (-5)^2 + 3^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$.
Now,$\vec{a} \cdot (\vec{v}_{1} \times \vec{v}_{2}) = (1)(-2) + (\lambda-2)(-5) + (2)(3) = -2 - 5\lambda + 10 + 6 = 14 - 5\lambda$.
Given $d = \frac{1}{\sqrt{38}}$,so $\frac{|14 - 5\lambda|}{\sqrt{38}} = \frac{1}{\sqrt{38}}$.
$|14 - 5\lambda| = 1$.
Case $1$: $14 - 5\lambda = 1 \Rightarrow 5\lambda = 13 \Rightarrow \lambda = 2.6$.
Case $2$: $14 - 5\lambda = -1 \Rightarrow 5\lambda = 15 \Rightarrow \lambda = 3$.
Since we need the integral value of $\lambda$,the answer is $3$.