Line Questions in English

(C) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are co-planar if the determinant of the vector connecting a point on each line and the direction vectors is zero:
$\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right| = 0$
Here,$(x_1, y_1, z_1) = (k, 2, 3)$ and $(x_2, y_2, z_2) = (-1, -2, -3)$.
The direction vectors are $(a_1, b_1, c_1) = (1, 2, 3)$ and $(a_2, b_2, c_2) = (3, 2, 1)$.
Substituting these values into the determinant:
$\left|\begin{array}{ccc} -1-k & -2-2 & -3-3 \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right| = 0$
$\left|\begin{array}{ccc} -(k+1) & -4 & -6 \\ 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right| = 0$
Expanding the determinant:
$-(k+1)(2-6) - (-4)(1-9) + (-6)(2-6) = 0$
$-(k+1)(-4) + 4(-8) - 6(-4) = 0$
$4(k+1) - 32 + 24 = 0$
$4k + 4 - 8 = 0$
$4k - 4 = 0$
$4k = 4$
$k = 1$

(A) The shortest distance 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 $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.
Here,$\vec{a}_1 = (1, 2, 3)$,$\vec{a}_2 = (2, 4, 5)$,$\vec{b}_1 = (2, 3, \lambda)$,and $\vec{b}_2 = (1, 4, 5)$.
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & \lambda \\ 1 & 4 & 5 \end{vmatrix} = \hat{i}(15-4\lambda) - \hat{j}(10-\lambda) + \hat{k}(8-3) = (15-4\lambda)\hat{i} + (\lambda-10)\hat{j} + 5\hat{k}$.
$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$.
$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(15-4\lambda) + 2(\lambda-10) + 2(5) = 15 - 4\lambda + 2\lambda - 20 + 10 = 5 - 2\lambda$.
$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(15-4\lambda)^2 + (\lambda-10)^2 + 25} = \sqrt{225 + 16\lambda^2 - 120\lambda + \lambda^2 - 20\lambda + 100 + 25} = \sqrt{17\lambda^2 - 140\lambda + 350}$.
Given $d = \frac{1}{\sqrt{3}}$,so $\frac{|5-2\lambda|}{\sqrt{17\lambda^2 - 140\lambda + 350}} = \frac{1}{\sqrt{3}}$.
Squaring both sides: $3(5-2\lambda)^2 = 17\lambda^2 - 140\lambda + 350$.
$3(25 - 20\lambda + 4\lambda^2) = 17\lambda^2 - 140\lambda + 350 \implies 75 - 60\lambda + 12\lambda^2 = 17\lambda^2 - 140\lambda + 350$.
$5\lambda^2 - 80\lambda + 275 = 0 \implies \lambda^2 - 16\lambda + 55 = 0$.
$(\lambda-5)(\lambda-11) = 0$,so $\lambda = 5$ or $\lambda = 11$.
The sum of all possible values of $\lambda$ is $5 + 11 = 16$.

(D) Let point $A$ on the first line be $(3\lambda+7, -\lambda+1, \lambda-2)$ and point $B$ on the second line be $(2\mu, 3\mu+7, \mu)$.
The direction ratios of line $AB$ are $(2\mu - (3\lambda+7), 3\mu+7 - (-\lambda+1), \mu - (\lambda-2))$,which simplifies to $(2\mu - 3\lambda - 7, 3\mu + \lambda + 6, \mu - \lambda + 2)$.
Since the direction ratios of the line $AB$ are given as $1, -4, 2$,we have:
$\frac{2\mu - 3\lambda - 7}{1} = \frac{3\mu + \lambda + 6}{-4} = \frac{\mu - \lambda + 2}{2} = k$
From the first and second ratios:
$-8\mu + 12\lambda + 28 = 3\mu + \lambda + 6 \implies 11\lambda - 11\mu = -22 \implies \lambda - \mu = -2 \implies \mu = \lambda + 2$.
From the second and third ratios:
$6\mu + 2\lambda + 12 = -4\mu + 4\lambda - 8 \implies 10\mu - 2\lambda = -20 \implies 5\mu - \lambda = -10$.
Substituting $\mu = \lambda + 2$ into $5\mu - \lambda = -10$:
$5(\lambda + 2) - \lambda = -10 \implies 4\lambda + 10 = -10 \implies 4\lambda = -20 \implies \lambda = -5$.
Then $\mu = -5 + 2 = -3$.
Coordinates of $A$: $(3(-5)+7, -(-5)+1, -5-2) = (-8, 6, -7)$.
Coordinates of $B$: $(2(-3), 3(-3)+7, -3) = (-6, -2, -3)$.
$(AB)^2 = (-6 - (-8))^2 + (-2 - 6)^2 + (-3 - (-7))^2 = (2)^2 + (-8)^2 + (4)^2 = 4 + 64 + 16 = 84$.

(B) 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 $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$.
Given $\vec{a}_1 = -\hat{i} + 3\hat{k}$,$\vec{b}_1 = \hat{i} - a\hat{j}$,$\vec{a}_2 = -\hat{j} + 2\hat{k}$,and $\vec{b}_2 = \hat{i} - \hat{j} + \hat{k}$.
$\vec{a}_2 - \vec{a}_1 = (0 - (-1))\hat{i} + (-1 - 0)\hat{j} + (2 - 3)\hat{k} = \hat{i} - \hat{j} - \hat{k}$.
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(-a) - \hat{j}(1) + \hat{k}(-1 + a) = -a\hat{i} - \hat{j} + (a-1)\hat{k}$.
$|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-a)^2 + (-1)^2 + (a-1)^2} = \sqrt{a^2 + 1 + a^2 - 2a + 1} = \sqrt{2a^2 - 2a + 2}$.
$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(-a) + (-1)(-1) + (-1)(a-1) = -a + 1 - a + 1 = 2 - 2a$.
Given $d = \sqrt{\frac{2}{3}}$,so $\frac{|2 - 2a|}{\sqrt{2a^2 - 2a + 2}} = \sqrt{\frac{2}{3}}$.
Squaring both sides: $\frac{4(1-a)^2}{2(a^2 - a + 1)} = \frac{2}{3} \implies \frac{2(1-2a+a^2)}{a^2-a+1} = \frac{2}{3} \implies 3(a^2 - 2a + 1) = a^2 - a + 1$.
$3a^2 - 6a + 3 = a^2 - a + 1 \implies 2a^2 - 5a + 2 = 0$.
$(2a - 1)(a - 2) = 0$,so $a = 2$ or $a = 0.5$.
The integral value of $a$ is $2$.

(B) The equation of line $l_{1}$ in the $xy$-plane $(z=0)$ with $x$-intercept $\frac{1}{8}$ and $y$-intercept $\frac{1}{4 \sqrt{2}}$ is given by $\frac{x}{1/8} + \frac{y}{1/(4 \sqrt{2})} = 1$,which simplifies to $8x + 4 \sqrt{2}y = 1$.
The direction vector of $l_{1}$ is $\vec{v}_{1} = (4 \sqrt{2}, -8, 0)$,which can be simplified to $(1, -\sqrt{2}, 0)$. The line passes through $A = (\frac{1}{8}, 0, 0)$.
The equation of line $l_{2}$ in the $zx$-plane $(y=0)$ with $x$-intercept $-\frac{1}{8}$ and $z$-intercept $-\frac{1}{6 \sqrt{3}}$ is given by $\frac{x}{-1/8} + \frac{z}{-1/(6 \sqrt{3})} = 1$,which simplifies to $-8x - 6 \sqrt{3}z = 1$.
The direction vector of $l_{2}$ is $\vec{v}_{2} = (-6 \sqrt{3}, 0, 8)$,which can be simplified to $(3 \sqrt{3}, 0, -4)$. The line passes through $B = (-\frac{1}{8}, 0, 0)$.
The shortest distance $d$ between two skew lines is given by $d = \frac{|(\vec{B}-\vec{A}) \cdot (\vec{v}_{1} \times \vec{v}_{2})|}{|\vec{v}_{1} \times \vec{v}_{2}|}$.
First,calculate $\vec{v}_{1} \times \vec{v}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -\sqrt{2} & 0 \\ 3 \sqrt{3} & 0 & -4 \end{vmatrix} = 4 \sqrt{2} \hat{i} + 4 \hat{j} + 3 \sqrt{6} \hat{k}$.
The magnitude $|\vec{v}_{1} \times \vec{v}_{2}| = \sqrt{(4 \sqrt{2})^2 + 4^2 + (3 \sqrt{6})^2} = \sqrt{32 + 16 + 54} = \sqrt{102}$.
$vec{B}-\vec{A} = (-\frac{1}{8} - \frac{1}{8}, 0, 0) = (-\frac{1}{4}, 0, 0)$.
$|(\vec{B}-\vec{A}) \cdot (\vec{v}_{1} \times \vec{v}_{2})| = |(-\frac{1}{4}, 0, 0) \cdot (4 \sqrt{2}, 4, 3 \sqrt{6})| = |-\sqrt{2}| = \sqrt{2}$.
$d = \frac{\sqrt{2}}{\sqrt{102}} = \frac{1}{\sqrt{51}}$.
Therefore,$d^{-2} = 51$.

(B) The line $l_{1}$ is given by $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-2}{0}$. The direction vector is $\vec{v_{1}} = (3, -2, 0)$.
The line $l_{2}$ is given by $\frac{x-1}{1}=\frac{y+3/2}{\alpha/2}=\frac{z+5}{2}$. The direction vector is $\vec{v_{2}} = (1, \alpha/2, 2)$.
Since $l_{1} \perp l_{2}$,their dot product is zero: $(3)(1) + (-2)(\alpha/2) + (0)(2) = 0$.
$3 - \alpha = 0 \Rightarrow \alpha = 3$.
The line $l_{3}$ is given by $\frac{x-1}{-3}=\frac{y-1/2}{-2}=\frac{z-0}{4}$. The direction vector is $\vec{v_{3}} = (-3, -2, 4)$.
For $\alpha = 3$,the direction vector of $l_{2}$ is $\vec{v_{2}} = (1, 3/2, 2)$.
The angle $\theta$ between $l_{2}$ and $l_{3}$ is given by $\cos \theta = \frac{|\vec{v_{2}} \cdot \vec{v_{3}}|}{||\vec{v_{2}}|| ||\vec{v_{3}}||}$.
$\vec{v_{2}} \cdot \vec{v_{3}} = (1)(-3) + (3/2)(-2) + (2)(4) = -3 - 3 + 8 = 2$.
$||\vec{v_{2}}|| = \sqrt{1^{2} + (3/2)^{2} + 2^{2}} = \sqrt{1 + 9/4 + 4} = \sqrt{29/4} = \frac{\sqrt{29}}{2}$.
$||\vec{v_{3}}|| = \sqrt{(-3)^{2} + (-2)^{2} + 4^{2}} = \sqrt{9 + 4 + 16} = \sqrt{29}$.
$\cos \theta = \frac{2}{(\sqrt{29}/2) \times \sqrt{29}} = \frac{2}{29/2} = \frac{4}{29}$.
Therefore,$\theta = \cos^{-1}\left(\frac{4}{29}\right) = \sec^{-1}\left(\frac{29}{4}\right)$.

(A) The given lines are $L_1: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $L_2: \frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$.
Points on the lines are $A(3, 2, 1)$ and $B(-3, 6, 5)$.
The direction vectors are $\vec{b_1} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b_2} = 2\hat{i} + \hat{j} + 3\hat{k}$.
The vector $\vec{AB} = (-3-3)\hat{i} + (6-2)\hat{j} + (5-1)\hat{k} = -6\hat{i} + 4\hat{j} + 4\hat{k}$.
The shortest distance $d$ is given by $d = \frac{|\vec{AB} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$.
First,calculate the cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3 \end{vmatrix} = \hat{i}(9 - (-1)) - \hat{j}(6 - (-2)) + \hat{k}(2 - 6) = 10\hat{i} - 8\hat{j} - 4\hat{k}$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{10^2 + (-8)^2 + (-4)^2} = \sqrt{100 + 64 + 16} = \sqrt{180} = 6\sqrt{5}$.
The scalar triple product $|\vec{AB} \cdot (\vec{b_1} \times \vec{b_2})| = |(-6)(10) + (4)(-8) + (4)(-4)| = |-60 - 32 - 16| = |-108| = 108$.
Thus,the shortest distance $d = \frac{108}{6\sqrt{5}} = \frac{18}{\sqrt{5}}$.

(D) Let $M$ be the foot of the perpendicular from $P(1, 2, 3)$ to the line $L$. Any point on the line $L$ is given by $M(3\lambda+6, 2\lambda+1, 3\lambda+2)$.
The direction vector of the line $L$ is $\vec{v} = 3\hat{i} + 2\hat{j} + 3\hat{k}$.
The vector $\vec{PM} = (3\lambda+6-1)\hat{i} + (2\lambda+1-2)\hat{j} + (3\lambda+2-3)\hat{k} = (3\lambda+5)\hat{i} + (2\lambda-1)\hat{j} + (3\lambda-1)\hat{k}$.
Since $\vec{PM} \perp \vec{v}$,their dot product is zero:
$3(3\lambda+5) + 2(2\lambda-1) + 3(3\lambda-1) = 0$
$9\lambda + 15 + 4\lambda - 2 + 9\lambda - 3 = 0$
$22\lambda + 10 = 0 \implies \lambda = -\frac{5}{11}$.
Substituting $\lambda$ into $M$,we get $M = (3(-\frac{5}{11})+6, 2(-\frac{5}{11})+1, 3(-\frac{5}{11})+2) = (\frac{51}{11}, \frac{1}{11}, \frac{7}{11})$.
Since $Q$ is the image of $P$ in the line,$M$ is the midpoint of $PQ$. Thus,$M = \frac{P+Q}{2} \implies Q = 2M - P = (2(\frac{51}{11})-1, 2(\frac{1}{11})-2, 2(\frac{7}{11})-3) = (\frac{91}{11}, -\frac{20}{11}, -\frac{19}{11})$.
$R$ divides $PQ$ in the ratio $1:3$,so $R = \frac{1(Q) + 3(P)}{1+3} = \frac{Q + 3P}{4}$.
$R = \frac{1}{4} ((\frac{91}{11} + 3), (-\frac{20}{11} + 6), (-\frac{19}{11} + 9)) = \frac{1}{4} (\frac{124}{11}, \frac{46}{11}, \frac{80}{11}) = (\frac{31}{11}, \frac{23}{22}, \frac{20}{11})$.
$\alpha+\beta+\gamma = \frac{62+23+40}{22} = \frac{125}{22}$.
$22(\alpha+\beta+\gamma) = 125$.

(A) The first line is $L_{1}: \frac{x+7}{-6}=\frac{y-6}{7}=\frac{z-0}{1}$.
$A$ point on $L_{1}$ is $\vec{a}_{1} = (-7, 6, 0)$ and its direction vector is $\vec{b}_{1} = (-6, 7, 1)$.
The second line is $L_{2}: \frac{x-7}{-2}=\frac{y-2}{1}=\frac{z-6}{1}$.
$A$ point on $L_{2}$ is $\vec{a}_{2} = (7, 2, 6)$ and its direction vector is $\vec{b}_{2} = (-2, 1, 1)$.
The vector $\vec{a}_{2} - \vec{a}_{1} = (7 - (-7), 2 - 6, 6 - 0) = (14, -4, 6)$.
The cross product $\vec{b}_{1} \times \vec{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -6 & 7 & 1 \\ -2 & 1 & 1 \end{vmatrix} = \hat{i}(7-1) - \hat{j}(-6+2) + \hat{k}(-6+14) = 6\hat{i} + 4\hat{j} + 8\hat{k} = (6, 4, 8)$.
The magnitude $|\vec{b}_{1} \times \vec{b}_{2}| = \sqrt{6^2 + 4^2 + 8^2} = \sqrt{36 + 16 + 64} = \sqrt{116} = 2\sqrt{29}$.
The shortest distance $d = \left| \frac{(\vec{a}_{2} - \vec{a}_{1}) \cdot (\vec{b}_{1} \times \vec{b}_{2})}{|\vec{b}_{1} \times \vec{b}_{2}|} \right| = \left| \frac{(14, -4, 6) \cdot (6, 4, 8)}{2\sqrt{29}} \right| = \left| \frac{84 - 16 + 48}{2\sqrt{29}} \right| = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}$.

(A) The direction ratios of the line are given by the cross product of the normals to the planes $x + y - z = 0$ and $x - 2y + 3z - 5 = 0$.
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & -2 & 3 \end{vmatrix} = \hat{i}(3 - 2) - \hat{j}(3 + 1) + \hat{k}(-2 - 1) = \hat{i} - 4\hat{j} - 3\hat{k}$.
The equation of the line passing through $(1, 2, 4)$ with direction vector $\vec{v} = (1, -4, -3)$ is $\vec{r} = (1, 2, 4) + \lambda(1, -4, -3)$.
Let $P$ be a point on the line: $P = (1 + \lambda, 2 - 4\lambda, 4 - 3\lambda)$.
Let $A = (1, -2, 5)$. The vector $\vec{AP} = P - A = (\lambda, 4 - 4\lambda, -1 - 3\lambda)$.
Since $\vec{AP}$ is perpendicular to the line direction $\vec{v} = (1, -4, -3)$,their dot product is zero:
$\vec{AP} \cdot \vec{v} = 1(\lambda) - 4(4 - 4\lambda) - 3(-1 - 3\lambda) = 0$.
$\lambda - 16 + 16\lambda + 3 + 9\lambda = 0 \implies 26\lambda - 13 = 0 \implies \lambda = \frac{1}{2}$.
Substituting $\lambda = \frac{1}{2}$ into $P$,we get $P = (1 + \frac{1}{2}, 2 - 2, 4 - \frac{3}{2}) = (\frac{3}{2}, 0, \frac{5}{2})$.
The length of the perpendicular is the distance $AP = \sqrt{(\frac{3}{2} - 1)^2 + (0 - (-2))^2 + (\frac{5}{2} - 5)^2} = \sqrt{(\frac{1}{2})^2 + (2)^2 + (-\frac{5}{2})^2} = \sqrt{\frac{1}{4} + 4 + \frac{25}{4}} = \sqrt{\frac{26}{4} + \frac{16}{4}} = \sqrt{\frac{42}{4}} = \sqrt{\frac{21}{2}}$.

(A) Let any point on the line be $A(\lambda) = (2\lambda - 1, 3\lambda - 2, 2\lambda + 1)$.
Given the distance from $P(4, 2, 7)$ is $\sqrt{26}$,we have:
$(2\lambda - 1 - 4)^2 + (3\lambda - 2 - 2)^2 + (2\lambda + 1 - 7)^2 = (\sqrt{26})^2$
$(2\lambda - 5)^2 + (3\lambda - 4)^2 + (2\lambda - 6)^2 = 26$
$(4\lambda^2 - 20\lambda + 25) + (9\lambda^2 - 24\lambda + 16) + (4\lambda^2 - 24\lambda + 36) = 26$
$17\lambda^2 - 68\lambda + 77 = 26$
$17\lambda^2 - 68\lambda + 51 = 0$
$\lambda^2 - 4\lambda + 3 = 0 \implies (\lambda - 1)(\lambda - 3) = 0$
So,$\lambda = 1$ and $\lambda = 3$.
For $\lambda = 1$,$Q = (1, 1, 3)$. For $\lambda = 3$,$R = (5, 7, 7)$.
$\overrightarrow{PQ} = Q - P = (1-4, 1-2, 3-7) = (-3, -1, -4)$.
$\overrightarrow{PR} = R - P = (5-4, 7-2, 7-7) = (1, 5, 0)$.
$\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -1 & -4 \\ 1 & 5 & 0 \end{vmatrix} = \hat{i}(0 - (-20)) - \hat{j}(0 - (-4)) + \hat{k}(-15 - (-1)) = 20\hat{i} - 4\hat{j} - 14\hat{k}$.
Area of $\triangle PQR = \frac{1}{2} |\overrightarrow{PQ} \times \overrightarrow{PR}| = \frac{1}{2} \sqrt{20^2 + (-4)^2 + (-14)^2} = \frac{1}{2} \sqrt{400 + 16 + 196} = \frac{1}{2} \sqrt{612} = \sqrt{\frac{612}{4}} = \sqrt{153}$.
The square of the area is $(\sqrt{153})^2 = 153$.

(B) Let the line be $L: \frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{-1} = \lambda$. Any point $M$ on the line is $(2\lambda-1, 3\lambda+3, -\lambda+1)$.
Vector $\vec{PM} = (2\lambda-1-a, 3\lambda+3-4, -\lambda+1-2) = (2\lambda-1-a, 3\lambda-1, -\lambda-1)$.
Since $\vec{PM}$ is perpendicular to the line direction $\vec{v} = (2, 3, -1)$,we have $\vec{PM} \cdot \vec{v} = 0$.
$2(2\lambda-1-a) + 3(3\lambda-1) - 1(-\lambda-1) = 0 \Rightarrow 4\lambda - 2 - 2a + 9\lambda - 3 + \lambda + 1 = 0 \Rightarrow 14\lambda - 4 - 2a = 0 \Rightarrow 7\lambda - 2 - a = 0 \Rightarrow a = 7\lambda - 2$.
Given the length of the perpendicular $PM = 2\sqrt{6}$,so $PM^2 = 24$.
$(2\lambda-1-a)^2 + (3\lambda-1)^2 + (-\lambda-1)^2 = 24$.
Substituting $a = 7\lambda-2$: $(2\lambda-1-(7\lambda-2))^2 + (3\lambda-1)^2 + (\lambda+1)^2 = 24$.
$(-5\lambda+1)^2 + (3\lambda-1)^2 + (\lambda+1)^2 = 24 \Rightarrow (25\lambda^2 - 10\lambda + 1) + (9\lambda^2 - 6\lambda + 1) + (\lambda^2 + 2\lambda + 1) = 24$.
$35\lambda^2 - 14\lambda + 3 = 24 \Rightarrow 35\lambda^2 - 14\lambda - 21 = 0 \Rightarrow 5\lambda^2 - 2\lambda - 3 = 0$.
$(5\lambda+3)(\lambda-1) = 0$. Since $a > 0$,$7\lambda-2 > 0 \Rightarrow \lambda > 2/7$. Thus $\lambda = 1$.
Then $a = 7(1)-2 = 5$. Point $M = (2(1)-1, 3(1)+3, -1+1) = (1, 6, 0)$.
Since $M$ is the midpoint of $PQ$,$\frac{a+\alpha_1}{2} = 1, \frac{4+\alpha_2}{2} = 6, \frac{2+\alpha_3}{2} = 0$.
$\alpha_1 = 2-5 = -3, \alpha_2 = 12-4 = 8, \alpha_3 = 0-2 = -2$.
$a + \alpha_1 + \alpha_2 + \alpha_3 = 5 - 3 + 8 - 2 = 8$.

(C) The given lines are $L_1: \frac{x-2}{3}=\frac{y+1}{2}=\frac{z-6}{2}$ and $L_2: \frac{x-6}{3}=\frac{y-1}{-2}=\frac{z+8}{0}$.
Here,point $A(2, -1, 6)$ lies on $L_1$ and point $B(6, 1, -8)$ lies on $L_2$.
The direction vectors are $\vec{b_1} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b_2} = 3\hat{i} - 2\hat{j} + 0\hat{k}$.
The vector $\vec{AB} = (6-2)\hat{i} + (1-(-1))\hat{j} + (-8-6)\hat{k} = 4\hat{i} + 2\hat{j} - 14\hat{k}$.
The cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 3 & -2 & 0 \end{vmatrix} = \hat{i}(0 - (-4)) - \hat{j}(0 - 6) + \hat{k}(-6 - 6) = 4\hat{i} + 6\hat{j} - 12\hat{k}$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{4^2 + 6^2 + (-12)^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$.
The shortest distance $d = \frac{|\vec{AB} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} = \frac{|(4)(4) + (2)(6) + (-14)(-12)|}{14} = \frac{|16 + 12 + 168|}{14} = \frac{196}{14} = 14$.

(D) The direction ratio of the line is given by the cross product of the normals to the two planes:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 0 & 3 & -1 \end{vmatrix} = \hat{i}(-2-3) - \hat{j}(-1-0) + \hat{k}(3-0) = -5\hat{i} + \hat{j} + 3\hat{k}$.
The equation of the line passing through $(3, 2, 1)$ with direction vector $\vec{v} = \langle -5, 1, 3 \rangle$ is $\frac{x-3}{-5} = \frac{y-2}{1} = \frac{z-1}{3} = \lambda$.
Any point $M$ on the line is given by $M(-5\lambda+3, \lambda+2, 3\lambda+1)$.
Let $P = (1, 9, 7)$. The vector $\vec{PM} = \langle -5\lambda+3-1, \lambda+2-9, 3\lambda+1-7 \rangle = \langle -5\lambda+2, \lambda-7, 3\lambda-6 \rangle$.
Since $\vec{PM}$ is perpendicular to the line,$\vec{PM} \cdot \langle -5, 1, 3 \rangle = 0$.
$-5(-5\lambda+2) + 1(\lambda-7) + 3(3\lambda-6) = 0$.
$25\lambda - 10 + \lambda - 7 + 9\lambda - 18 = 0$.
$35\lambda - 35 = 0 \implies \lambda = 1$.
Substituting $\lambda = 1$ into the coordinates of $M$,we get $M = (-5(1)+3, 1+2, 3(1)+1) = (-2, 3, 4)$.
Thus,$(\alpha, \beta, \gamma) = (-2, 3, 4)$.
Therefore,$\alpha+\beta+\gamma = -2+3+4 = 5$.

(D) The lines are $L_1: \frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$ and $L_2: \frac{x-\lambda}{3}=\frac{y-2\sqrt{6}}{4}=\frac{z+2\sqrt{6}}{5}$.
Points on the lines are $P_1(-\sqrt{6}, \sqrt{6}, \sqrt{6})$ and $P_2(\lambda, 2\sqrt{6}, -2\sqrt{6})$.
Direction vectors are $\vec{v_1} = (2, 3, 4)$ and $\vec{v_2} = (3, 4, 5)$.
The cross product is $\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}$.
The magnitude is $|\vec{n}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}$.
The shortest distance is $d = \frac{|(\vec{P_2} - \vec{P_1}) \cdot \vec{n}|}{|\vec{n}|} = 6$.
$\vec{P_2} - \vec{P_1} = (\lambda + \sqrt{6}, \sqrt{6}, -3\sqrt{6})$.
$|(\lambda + \sqrt{6})(-1) + (\sqrt{6})(2) + (-3\sqrt{6})(-1)| = 6 \times \sqrt{6} = 6\sqrt{6}$.
$|-\lambda - \sqrt{6} + 2\sqrt{6} + 3\sqrt{6}| = 6\sqrt{6} \Rightarrow |4\sqrt{6} - \lambda| = 6\sqrt{6}$.
Case $1: 4\sqrt{6} - \lambda = 6\sqrt{6} \Rightarrow \lambda = -2\sqrt{6}$.
Case $2: 4\sqrt{6} - \lambda = -6\sqrt{6} \Rightarrow \lambda = 10\sqrt{6}$.
Sum of values $= -2\sqrt{6} + 10\sqrt{6} = 8\sqrt{6}$.
Square of the sum $= (8\sqrt{6})^2 = 64 \times 6 = 384$.

(A) Let the point $P$ on $L_1$ be $(2\lambda+1, \lambda+3, 2\lambda+2)$ and the point $Q$ on $L_2$ be $(\mu+2, 2\mu+2, 3\mu+3)$.
The direction ratios of the line $PQ$ are given by $((\mu+2)-(2\lambda+1), (2\mu+2)-(\lambda+3), (3\mu+3)-(2\lambda+2)) = (\mu-2\lambda+1, 2\mu-\lambda-1, 3\mu-2\lambda+1)$.
Since the direction ratios of $L_3$ are $1, -1, -2$,we have:
$\frac{\mu-2\lambda+1}{1} = \frac{2\mu-\lambda-1}{-1} = \frac{3\mu-2\lambda+1}{-2}$.
From the first two parts: $-\mu+2\lambda-1 = 2\mu-\lambda-1 \Rightarrow 3\lambda = 3\mu \Rightarrow \lambda = \mu$.
Substituting $\lambda = \mu$ into the equality $\frac{\mu-2\lambda+1}{1} = \frac{3\mu-2\lambda+1}{-2}$:
$\frac{-\lambda+1}{1} = \frac{\lambda+1}{-2} \Rightarrow 2\lambda - 2 = \lambda + 1 \Rightarrow \lambda = 3$.
Thus,$\lambda = 3$ and $\mu = 3$. The points are $P(7, 6, 8)$ and $Q(5, 8, 12)$.
The length $PQ = \sqrt{(5-7)^2 + (8-6)^2 + (12-8)^2} = \sqrt{(-2)^2 + 2^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.

(D) The equation of the line passing through $(-3, 2, 3)$ with direction ratios $(3, 3, -1)$ is given by $\frac{x+3}{3} = \frac{y-2}{3} = \frac{z-3}{-1} = \lambda$.
Any point $M$ on this line is $(3\lambda-3, 3\lambda+2, 3-\lambda)$.
The direction ratios of the vector $\vec{PM}$ are $(3\lambda-3-4, 3\lambda+2-6, 3-\lambda-(-2)) = (3\lambda-7, 3\lambda-4, 5-\lambda)$.
Since $\vec{PM}$ is perpendicular to the line with direction ratios $(3, 3, -1)$,their dot product must be zero:
$3(3\lambda-7) + 3(3\lambda-4) - 1(5-\lambda) = 0$.
$9\lambda - 21 + 9\lambda - 12 - 5 + \lambda = 0$.
$19\lambda - 38 = 0 \Rightarrow \lambda = 2$.
Substituting $\lambda = 2$ into the coordinates of $M$,we get $M(3(2)-3, 3(2)+2, 3-2) = (3, 8, 1)$.
The distance $PM = \sqrt{(3-4)^2 + (8-6)^2 + (1-(-2))^2} = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1+4+9} = \sqrt{14}$.

(A) First,express the lines in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
For the first line: $x+1 = 2y = -12z \Rightarrow \frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12}$. Point $A = (-1, 0, 0)$,direction vector $\vec{p} = (1, 1/2, -1/12)$.
For the second line: $x = y+2 = 6z-6 \Rightarrow \frac{x}{1} = \frac{y+2}{1} = \frac{z-1}{1/6}$. Point $B = (0, -2, 1)$,direction vector $\vec{q} = (1, 1, 1/6)$.
The vector $\vec{B}-\vec{A} = (0-(-1), -2-0, 1-0) = (1, -2, 1) = \hat{i} - 2\hat{j} + \hat{k}$.
The cross product $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1/2 & -1/12 \\ 1 & 1 & 1/6 \end{vmatrix} = \hat{i}(\frac{1}{12} + \frac{1}{12}) - \hat{j}(\frac{1}{6} + \frac{1}{12}) + \hat{k}(1 - \frac{1}{2}) = \frac{1}{6}\hat{i} - \frac{1}{4}\hat{j} + \frac{1}{2}\hat{k}$.
Scaling the vector $\vec{p} \times \vec{q}$ by $12$,we get $2\hat{i} - 3\hat{j} + 6\hat{k}$.
The magnitude $|\vec{p} \times \vec{q}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$.
The shortest distance is $\left| \frac{(\vec{B}-\vec{A}) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right| = \left| \frac{(1, -2, 1) \cdot (2, -3, 6)}{7} \right| = \left| \frac{2 + 6 + 6}{7} \right| = \frac{14}{7} = 2$.

(C) Let the line be $L: \frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1} = \lambda$.
Any point on the line is $P(2\lambda - 1, 5\lambda + 1, -\lambda - 1)$.
Since $P$ is the foot of the perpendicular from $A(2, 0, 5)$ to the line,the vector $\vec{AP}$ must be perpendicular to the direction vector of the line $\vec{b} = 2\hat{i} + 5\hat{j} - \hat{k}$.
$\vec{AP} = (2\lambda - 1 - 2)\hat{i} + (5\lambda + 1 - 0)\hat{j} + (-\lambda - 1 - 5)\hat{k} = (2\lambda - 3)\hat{i} + (5\lambda + 1)\hat{j} + (-\lambda - 6)\hat{k}$.
Since $\vec{AP} \cdot \vec{b} = 0$,we have:
$2(2\lambda - 3) + 5(5\lambda + 1) - 1(-\lambda - 6) = 0$
$4\lambda - 6 + 25\lambda + 5 + \lambda + 6 = 0$
$30\lambda + 5 = 0 \Rightarrow \lambda = -\frac{1}{6}$.
Substituting $\lambda = -\frac{1}{6}$ into the coordinates of $P$:
$\alpha = 2(-\frac{1}{6}) - 1 = -\frac{1}{3} - 1 = -\frac{4}{3}$
$\beta = 5(-\frac{1}{6}) + 1 = -\frac{5}{6} + 1 = \frac{1}{6}$
$\gamma = -(-\frac{1}{6}) - 1 = \frac{1}{6} - 1 = -\frac{5}{6}$
Now check the options:
$A) \frac{\alpha \beta}{\gamma} = \frac{(-\frac{4}{3})(\frac{1}{6})}{-\frac{5}{6}} = \frac{-\frac{4}{18}}{-\frac{5}{6}} = \frac{4}{18} \times \frac{6}{5} = \frac{4}{15}$ (Correct)
$B) \frac{\alpha}{\beta} = \frac{-\frac{4}{3}}{\frac{1}{6}} = -\frac{4}{3} \times 6 = -8$ (Correct)
$C) \frac{\beta}{\gamma} = \frac{\frac{1}{6}}{-\frac{5}{6}} = -\frac{1}{5}$ (Incorrect,as it is given as $-5$)
$D) \frac{\gamma}{\alpha} = \frac{-\frac{5}{6}}{-\frac{4}{3}} = \frac{5}{6} \times \frac{3}{4} = \frac{5}{8}$ (Correct)
Thus,option $C$ is not correct.

(A) The line joining $(1, 2, 3)$ and $(2, 3, 4)$ has the direction vector $\vec{p} = (2-1)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k} = \hat{i} + \hat{j} + \hat{k}$. The equation of this line is $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} + \hat{k})$.
The second line is $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} - \hat{j} + 0\hat{k})$.
Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$,$\vec{p} = \hat{i} + \hat{j} + \hat{k}$,and $\vec{q} = 2\hat{i} - \hat{j}$.
The cross product $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} = \hat{i}(0 - (-1)) - \hat{j}(0 - 2) + \hat{k}(-1 - 2) = \hat{i} + 2\hat{j} - 3\hat{k}$.
The magnitude $|\vec{p} \times \vec{q}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
The vector $\vec{b} - \vec{a} = (1-1)\hat{i} + (-1-2)\hat{j} + (2-3)\hat{k} = -3\hat{j} - \hat{k}$.
The shortest distance $\alpha = \left| \frac{(\vec{b} - \vec{a}) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right| = \left| \frac{(0\hat{i} - 3\hat{j} - 1\hat{k}) \cdot (1\hat{i} + 2\hat{j} - 3\hat{k})}{\sqrt{14}} \right| = \left| \frac{0 - 6 + 3}{\sqrt{14}} \right| = \frac{3}{\sqrt{14}}$.
Thus,$28\alpha^2 = 28 \times \left( \frac{3}{\sqrt{14}} \right)^2 = 28 \times \frac{9}{14} = 2 \times 9 = 18$.

(B) The given lines are $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$ and $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$.
For the first line, a point $\vec{a} = \hat{i} - 8\hat{j} + 4\hat{k}$ and direction vector $\vec{p} = 2\hat{i} - 7\hat{j} + 5\hat{k}$.
For the second line, a point $\vec{b} = \hat{i} + 2\hat{j} + 6\hat{k}$ and direction vector $\vec{q} = 2\hat{i} + \hat{j} - 3\hat{k}$.
The cross product $\vec{p} \times \vec{q}$ is given by:
$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21-5) - \hat{j}(-6-10) + \hat{k}(2+14) = 16\hat{i} + 16\hat{j} + 16\hat{k} = 16(\hat{i} + \hat{j} + \hat{k})$.
The magnitude is $|\vec{p} \times \vec{q}| = 16\sqrt{1^2 + 1^2 + 1^2} = 16\sqrt{3}$.
The vector $(\vec{a} - \vec{b}) = (1-1)\hat{i} + (-8-2)\hat{j} + (4-6)\hat{k} = -10\hat{j} - 2\hat{k}$.
The shortest distance $d$ is given by $d = \left| \frac{(\vec{a} - \vec{b}) \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right|$.
$d = \left| \frac{(-10\hat{j} - 2\hat{k}) \cdot 16(\hat{i} + \hat{j} + \hat{k})}{16\sqrt{3}} \right| = \left| \frac{16(0 - 10 - 2)}{16\sqrt{3}} \right| = \left| \frac{-12}{\sqrt{3}} \right| = \frac{12}{\sqrt{3}} = 4\sqrt{3}$.

(C) The line $l_1$ passes through $A(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. The direction vector of the plane is $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$. Thus,the equation of line $l_1$ is $\frac{x - 2}{2} = \frac{y - 6}{1} = \frac{z - 2}{-2}$.
The second line is $l_2: \frac{x + 1}{2} = \frac{y + 4}{-3} = \frac{z}{2}$,which passes through $B(-1, -4, 0)$ with direction vector $\vec{v_2} = 2\hat{i} - 3\hat{j} + 2\hat{k}$.
The shortest distance $d$ between two skew lines is given by $d = \left| \frac{(\vec{b_2} - \vec{b_1}) \cdot (\vec{v_1} \times \vec{v_2})}{|\vec{v_1} \times \vec{v_2}|} \right|$.
Here,$\vec{b_1} = 2\hat{i} + 6\hat{j} + 2\hat{k}$ and $\vec{b_2} = -\hat{i} - 4\hat{j} + 0\hat{k}$.
$\vec{b_2} - \vec{b_1} = (-1 - 2)\hat{i} + (-4 - 6)\hat{j} + (0 - 2)\hat{k} = -3\hat{i} - 10\hat{j} - 2\hat{k}$.
$\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{vmatrix} = \hat{i}(2 - 6) - \hat{j}(4 - (-4)) + \hat{k}(-6 - 2) = -4\hat{i} - 8\hat{j} - 8\hat{k}$.
$|\vec{v_1} \times \vec{v_2}| = \sqrt{(-4)^2 + (-8)^2 + (-8)^2} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12$.
$d = \left| \frac{(-3\hat{i} - 10\hat{j} - 2\hat{k}) \cdot (-4\hat{i} - 8\hat{j} - 8\hat{k})}{12} \right| = \left| \frac{12 + 80 + 16}{12} \right| = \frac{108}{12} = 9$.

(B) The direction vector $\vec{v}$ of the line is given by the cross product of the normals to the planes $x + 3y - 2z - 2 = 0$ and $x - y + 2z = 0$.
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ 1 & -1 & 2 \end{vmatrix} = \hat{i}(6 - 2) - \hat{j}(2 + 2) + \hat{k}(-1 - 3) = 4\hat{i} - 4\hat{j} - 4\hat{k}$.
We can take the direction vector as $\vec{d} = (1, -1, -1)$.
The equation of line $L$ passing through $P(2, 3, 1)$ is $\frac{x - 2}{1} = \frac{y - 3}{-1} = \frac{z - 1}{-1} = k$.
Any point $R$ on the line is $(k + 2, -k + 3, -k + 1)$.
Let $Q = (5, 3, 8)$. The vector $\vec{QR} = (k + 2 - 5, -k + 3 - 3, -k + 1 - 8) = (k - 3, -k, -k - 7)$.
Since $\vec{QR}$ is perpendicular to the line direction $(1, -1, -1)$,we have:
$1(k - 3) - 1(-k) - 1(-k - 7) = 0 \Rightarrow k - 3 + k + k + 7 = 0 \Rightarrow 3k + 4 = 0 \Rightarrow k = -\frac{4}{3}$.
The vector $\vec{QR} = (-\frac{4}{3} - 3, -(-\frac{4}{3}), -(-\frac{4}{3}) - 7) = (-\frac{13}{3}, \frac{4}{3}, -\frac{17}{3})$.
The distance $\alpha = |\vec{QR}| = \sqrt{(-\frac{13}{3})^2 + (\frac{4}{3})^2 + (-\frac{17}{3})^2} = \sqrt{\frac{169 + 16 + 289}{9}} = \sqrt{\frac{474}{9}}$.
Thus,$\alpha^2 = \frac{474}{9}$.
Therefore,$3\alpha^2 = 3 \times \frac{474}{9} = \frac{474}{3} = 158$.

(A) The line $L$ passes through $(5, \lambda, -\lambda)$ with direction vector $\vec{b_1} = (-2, 0, 1)$.
The line $L_1$ can be written as $\frac{x+1}{1} = \frac{y-1}{1} = \frac{z-4}{-1}$,passing through $(-1, 1, 4)$ with direction vector $\vec{b_2} = (1, 1, -1)$.
The cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & 1 \\ 1 & 1 & -1 \end{vmatrix} = -\hat{i} - \hat{j} - 2\hat{k}$.
The shortest distance 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 $\vec{a_2} - \vec{a_1} = (-1-5, 1-\lambda, 4-(-\lambda)) = (-6, 1-\lambda, 4+\lambda)$.
$d = \frac{|(-6)(-1) + (1-\lambda)(-1) + (4+\lambda)(-2)|}{\sqrt{(-1)^2 + (-1)^2 + (-2)^2}} = \frac{|6 - 1 + \lambda - 8 - 2\lambda|}{\sqrt{6}} = \frac{|-\lambda - 3|}{\sqrt{6}}$.
Given $d = 2\sqrt{6}$,so $|\lambda+3| = 12$. Since $\lambda \geq 0$,$\lambda = 9$.
For $L$,$(\alpha, \beta, \gamma) = (5-2k, \lambda, k-\lambda) = (5-2k, 9, k-9)$.
Then $\alpha = 5-2k$ and $\gamma = k-9$. Thus $k = \gamma+9$.
Substituting $k$: $\alpha = 5-2(\gamma+9) = 5-2\gamma-18 = -2\gamma-13$,so $\alpha+2\gamma = -13$.
Checking options: $\alpha+2\gamma=24$ is not possible.

(C) The shortest distance $d$ between two lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{a_2}=\frac{z-z_1}{a_3}$ and $\frac{x-x_2}{b_1}=\frac{y-y_2}{b_2}=\frac{z-z_2}{b_3}$ is given by the formula:
$d = \frac{|(\vec{r_2} - \vec{r_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$
Here, $\vec{r_1} = (5, 2, 4)$, $\vec{r_2} = (-3, -5, 1)$, $\vec{b_1} = (1, 2, -3)$, and $\vec{b_2} = (1, 4, -5)$.
First, calculate $\vec{r_2} - \vec{r_1} = (-3-5, -5-2, 1-4) = (-8, -7, -3)$.
Next, calculate the cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix} = \hat{i}(-10+12) - \hat{j}(-5+3) + \hat{k}(4-2) = 2\hat{i} + 2\hat{j} + 2\hat{k}$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}$.
The dot product $(\vec{r_2} - \vec{r_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (-8)(2) + (-7)(2) + (-3)(2) = -16 - 14 - 6 = -36$.
Thus, $d = \frac{|-36|}{2\sqrt{3}} = \frac{36}{2\sqrt{3}} = \frac{18}{\sqrt{3}} = 6\sqrt{3}$.

(D) The diagonal $OP$ passes through $(0, 0, 0)$ and $(3, 4, 5)$. Its direction vector is $\vec{b}_1 = 3\hat{i} + 4\hat{j} + 5\hat{k}$. The equation of $OP$ is $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$.
An edge parallel to the $z$-axis not passing through $O(0, 0, 0)$ or $P(3, 4, 5)$ must pass through the vertex $(3, 0, 0)$ or $(0, 4, 0)$. Let us consider the edge passing through $(3, 0, 0)$. Its direction vector is $\vec{b}_2 = \hat{k} = (0, 0, 1)$.
The shortest distance $d$ between two skew lines $\vec{r} = \vec{a}_1 + t\vec{b}_1$ and $\vec{r} = \vec{a}_2 + s\vec{b}_2$ 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,$\vec{a}_1 = (0, 0, 0)$,$\vec{a}_2 = (3, 0, 0)$,$\vec{b}_1 = (3, 4, 5)$,and $\vec{b}_2 = (0, 0, 1)$.
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 0 & 0 & 1 \end{vmatrix} = 4\hat{i} - 3\hat{j}$.
$|\vec{b}_1 \times \vec{b}_2| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5$.
$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (3, 0, 0) \cdot (4, -3, 0) = 12$.
Therefore,$d = \frac{|12|}{5} = \frac{12}{5}$.

(A) The shortest distance $S_d$ between two skew lines $\vec{r} = \vec{a} + \lambda \vec{n}_1$ and $\vec{r} = \vec{b} + \mu \vec{n}_2$ is given by the formula:
$S_d = \left| \frac{(\vec{b} - \vec{a}) \cdot (\vec{n}_1 \times \vec{n}_2)}{|\vec{n}_1 \times \vec{n}_2|} \right|$
From the given equations:
$\vec{a} = (4, -2, -3)$,$\vec{b} = (1, 3, 4)$
$\vec{n}_1 = (4, 5, 3)$,$\vec{n}_2 = (3, 4, 2)$
First,calculate the cross product $\vec{n}_1 \times \vec{n}_2$:
$\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & 3 \\ 3 & 4 & 2 \end{vmatrix} = \hat{i}(10-12) - \hat{j}(8-9) + \hat{k}(16-15) = -2\hat{i} + \hat{j} + \hat{k} = (-2, 1, 1)$
The magnitude is $|\vec{n}_1 \times \vec{n}_2| = \sqrt{(-2)^2 + 1^2 + 1^2} = \sqrt{4+1+1} = \sqrt{6}$
Now,calculate $\vec{b} - \vec{a} = (1-4, 3-(-2), 4-(-3)) = (-3, 5, 7)$
The dot product $(\vec{b} - \vec{a}) \cdot (\vec{n}_1 \times \vec{n}_2) = (-3, 5, 7) \cdot (-2, 1, 1) = 6 + 5 + 7 = 18$
Therefore,$S_d = \left| \frac{18}{\sqrt{6}} \right| = \frac{18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{18\sqrt{6}}{6} = 3\sqrt{6}$

(B) The given lines are $L_1: \frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$ and $L_2: \frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$.
From $L_1$,a point $P_1 = (-2, 0, 5)$ and direction vector $\vec{b_1} = \hat{i} - 2\hat{j} + 2\hat{k}$.
From $L_2$,a point $P_2 = (4, 1, -3)$ and direction vector $\vec{b_2} = \hat{i} + 2\hat{j} + 0\hat{k}$.
The vector $\vec{P_1P_2} = (4 - (-2))\hat{i} + (1 - 0)\hat{j} + (-3 - 5)\hat{k} = 6\hat{i} + \hat{j} - 8\hat{k}$.
The cross product $\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(2 - (-2)) = -4\hat{i} + 2\hat{j} + 4\hat{k}$.
The magnitude $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-4)^2 + 2^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.
The shortest distance $d = \frac{|\vec{P_1P_2} \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} = \frac{|(6\hat{i} + \hat{j} - 8\hat{k}) \cdot (-4\hat{i} + 2\hat{j} + 4\hat{k})|}{6} = \frac{|-24 + 2 - 32|}{6} = \frac{|-54|}{6} = 9$.

(B) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar if $\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right| = 0$.
Given line $L_1: \frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5}$,point $P_1(-3, 1, 5)$ and direction vector $\vec{v_1} = (-3, 1, 5)$.
For option $B$: Line $L_2: \frac{x+1}{-1} = \frac{y-2}{2} = \frac{z-5}{5}$,point $P_2(-1, 2, 5)$ and direction vector $\vec{v_2} = (-1, 2, 5)$.
Vector $\vec{P_1P_2} = (-1 - (-3), 2 - 1, 5 - 5) = (2, 1, 0)$.
Check coplanarity condition:
$\left|\begin{array}{ccc}2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5\end{array}\right| = 2(5 - 10) - 1(-15 - (-5)) + 0 = 2(-5) - 1(-10) = -10 + 10 = 0$.
Since the determinant is $0$,the lines are coplanar. Thus,option $B$ is correct.

(C) The lines are $L_1: \frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}$ and $L_2: \frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}$.
The points on the lines are $A(\lambda, 3, -6)$ and $B(-\lambda, 0, 6)$. The vector $\vec{AB} = (-2\lambda, -3, 12)$.
The direction vectors are $\vec{v_1} = (0, 4, 1)$ and $\vec{v_2} = (3, -4, 0)$.
The cross product $\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 4 & 1 \\ 3 & -4 & 0 \end{vmatrix} = \hat{i}(4) - \hat{j}(-3) + \hat{k}(-12) = 4\hat{i} + 3\hat{j} - 12\hat{k}$.
The magnitude $|\vec{v_1} \times \vec{v_2}| = \sqrt{4^2 + 3^2 + (-12)^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$.
The shortest distance $d = \frac{|\vec{AB} \cdot (\vec{v_1} \times \vec{v_2})|}{|\vec{v_1} \times \vec{v_2}|} = 13$.
$|(-2\lambda, -3, 12) \cdot (4, 3, -12)| = 13 \times 13 = 169$.
$|-8\lambda - 9 - 144| = 169 \implies |-8\lambda - 153| = 169$.
$8\lambda + 153 = 169$ or $8\lambda + 153 = -169$.
$8\lambda = 16 \implies \lambda_1 = 2$.
$8\lambda = -322 \implies \lambda_2 = -\frac{322}{8} = -40.25$.
The sum $\sum_{\lambda \in S} \lambda = 2 - \frac{322}{8} = \frac{16 - 322}{8} = -\frac{306}{8}$.
Then $8\left|\sum_{\lambda \in S} \lambda\right| = 8 \times \left|-\frac{306}{8}\right| = 306$.

(B) Let the line along which the distance is measured be $\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6} = \lambda$.
Any point on this line is $P(2\lambda + 7, -3\lambda - 2, 6\lambda + 11)$.
Since this point $P$ also lies on the line $\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$,we have:
$\frac{2\lambda + 7 - 6}{1} = \frac{-3\lambda - 2 - 4}{0} = \frac{6\lambda + 11 - 8}{3}$
From the middle term,the denominator is $0$,so the numerator must be $0$ for the ratio to be defined:
$-3\lambda - 6 = 0 \Rightarrow \lambda = -2$.
Substituting $\lambda = -2$ into the coordinates of $P$:
$x = 2(-2) + 7 = 3$
$y = -3(-2) - 2 = 4$
$z = 6(-2) + 11 = -1$
So,the point of intersection is $B(3, 4, -1)$.
The distance $AB$ is the distance between $A(7, -2, 11)$ and $B(3, 4, -1)$:
$AB = \sqrt{(7-3)^2 + (-2-4)^2 + (11 - (-1))^2}$
$AB = \sqrt{4^2 + (-6)^2 + 12^2}$
$AB = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$.

(B) The given lines are $L_1: \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $L_2: \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$.
The points on the lines are $A(4, -1, 0)$ and $B(\lambda, -1, 2)$. The direction vectors are $\vec{d_1} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{d_2} = 2\hat{i} + 4\hat{j} - 5\hat{k}$.
The shortest distance $d$ is given by $d = \left|\frac{(\vec{b}-\vec{a}) \cdot (\vec{d_1} \times \vec{d_2})}{|\vec{d_1} \times \vec{d_2}|}\right|$.
First,calculate $\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10+12) - \hat{j}(-5+6) + \hat{k}(4-4) = 2\hat{i} - \hat{j} + 0\hat{k}$.
The magnitude is $|\vec{d_1} \times \vec{d_2}| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}$.
Now,$\vec{b}-\vec{a} = (\lambda-4)\hat{i} + 0\hat{j} + 2\hat{k}$.
The dot product is $(\vec{b}-\vec{a}) \cdot (\vec{d_1} \times \vec{d_2}) = (\lambda-4)(2) + 0(-1) + 2(0) = 2(\lambda-4)$.
Given the distance is $\frac{6}{\sqrt{5}}$,we have $\frac{|2(\lambda-4)|}{\sqrt{5}} = \frac{6}{\sqrt{5}}$.
This simplifies to $|2(\lambda-4)| = 6$,or $|\lambda-4| = 3$.
Thus,$\lambda-4 = 3$ or $\lambda-4 = -3$,which gives $\lambda = 7$ or $\lambda = 1$.
The sum of all possible values of $\lambda$ is $7 + 1 = 8$.

(C) Let the line be $L_1: \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$. Any point $M$ on the line is $(\lambda, 1+2\lambda, 2+3\lambda)$.
Let $P = (1,0,7)$. The vector $\overrightarrow{PM} = (\lambda-1, 1+2\lambda, 3\lambda-5)$.
Since $\overrightarrow{PM}$ is perpendicular to the line $L_1$ with direction vector $\vec{b} = (1,2,3)$,we have $\overrightarrow{PM} \cdot \vec{b} = 0$.
$(\lambda-1)(1) + (1+2\lambda)(2) + (3\lambda-5)(3) = 0 \Rightarrow \lambda-1+2+4\lambda+9\lambda-15 = 0 \Rightarrow 14\lambda = 14 \Rightarrow \lambda = 1$.
Thus,$M = (1, 3, 5)$.
Since $M$ is the midpoint of $PQ$,where $Q = (\alpha, \beta, \gamma)$,we have $M = \frac{P+Q}{2} \Rightarrow Q = 2M - P = 2(1,3,5) - (1,0,7) = (2-1, 6-0, 10-7) = (1,6,3)$.
So,$(\alpha, \beta, \gamma) = (1,6,3)$.
Let the direction cosines of the required line be $(l, m, n)$. Given $m = \cos(\frac{2\pi}{3}) = -\frac{1}{2}$ and $n = \cos(\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}}$.
Since $l^2+m^2+n^2=1$,$l^2 + (-\frac{1}{2})^2 + (-\frac{1}{\sqrt{2}})^2 = 1 \Rightarrow l^2 + \frac{1}{4} + \frac{1}{2} = 1 \Rightarrow l^2 = \frac{1}{4}$.
Since the line makes an acute angle with the $x$-axis,$l = \frac{1}{2}$.
The line passes through $(1,6,3)$ with direction vector $(\frac{1}{2}, -\frac{1}{2}, -\frac{1}{\sqrt{2}})$,or $(1, -1, -\sqrt{2})$.
The equation is $\frac{x-1}{1} = \frac{y-6}{-1} = \frac{z-3}{-\sqrt{2}} = \mu$.
For $\mu = 2$,$x = 3, y = 4, z = 3-2\sqrt{2}$. This matches option $C$.

(A) Let the first line be $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16} = \lambda$. Then any point on this line is $(2\lambda+2, -2\lambda, 16\lambda+7)$.
Let the second line be $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1} = k$. Then any point on this line is $(4k-3, 3k-2, k-2)$.
Since they intersect at $P$,we equate the coordinates:
$2\lambda+2 = 4k-3 \Rightarrow 2\lambda - 4k = -5$
$-2\lambda = 3k-2 \Rightarrow 2\lambda + 3k = 2$
Adding these equations: $-k = -7 \Rightarrow k=1$. Substituting $k=1$ in $2\lambda+3(1)=2$,we get $2\lambda = -1 \Rightarrow \lambda = -1/2$.
Checking the $z$-coordinate: $16(-1/2)+7 = -8+7 = -1$ and $k-2 = 1-2 = -1$. Since they match,the intersection point $P$ is $(4(1)-3, 3(1)-2, 1-2) = (1, 1, -1)$.
Now,we find the distance $l$ of $P(1, 1, -1)$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$.
Let $A = (-1, 1, 1)$ be a point on the line and $\vec{v} = 2\hat{i}+3\hat{j}+\hat{k}$ be the direction vector.
Vector $\vec{AP} = (1 - (-1))\hat{i} + (1-1)\hat{j} + (-1-1)\hat{k} = 2\hat{i} - 2\hat{k}$.
The distance $l$ is given by $l = \frac{|\vec{AP} \times \vec{v}|}{|\vec{v}|}$.
$\vec{AP} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(0 - (-6)) - \hat{j}(2 - (-4)) + \hat{k}(6 - 0) = 6\hat{i} - 6\hat{j} + 6\hat{k}$.
$|\vec{AP} \times \vec{v}| = \sqrt{6^2 + (-6)^2 + 6^2} = \sqrt{36+36+36} = \sqrt{108} = 6\sqrt{3}$.
$|\vec{v}| = \sqrt{2^2+3^2+1^2} = \sqrt{4+9+1} = \sqrt{14}$.
$l = \frac{6\sqrt{3}}{\sqrt{14}} \Rightarrow l^2 = \frac{36 \times 3}{14} = \frac{108}{14}$.
Therefore,$14l^2 = 108$.

(B) Let $P$ be a point on the line $x = y + 2 = z = t$. Then $P = (t, t - 2, t)$.
Let $Q$ be a point on the line $x + 2 = 2y = 2z = 2s$. Then $Q = (2s - 2, s, s)$.
The direction ratios of the line $PQ$ are given as $(2, 1, 2)$.
Thus,the direction ratios of the vector $\vec{PQ}$ are $(2s - 2 - t, s - (t - 2), s - t) = (2s - t - 2, s - t + 2, s - t)$.
Since the line $PQ$ has direction ratios $(2, 1, 2)$,we have:
$\frac{2s - t - 2}{2} = \frac{s - t + 2}{1} = \frac{s - t}{2} = k$ (say).
From $\frac{s - t + 2}{1} = \frac{s - t}{2}$,we get $2s - 2t + 4 = s - t$,so $s - t = -4$.
Substituting $s - t = -4$ into the ratios:
$\frac{2s - t - 2}{2} = \frac{-4 + 2}{1} = -2$,so $2s - t - 2 = -4$,which means $2s - t = -2$.
Solving $s - t = -4$ and $2s - t = -2$,we get $s = 2$ and $t = 6$.
Thus,$P = (6, 4, 6)$ and $Q = (2, 2, 2)$.
The equation of line $PQ$ is $\frac{x - 2}{2} = \frac{y - 2}{1} = \frac{z - 2}{2} = \lambda$.
Any point $F$ on $PQ$ is $(2\lambda + 2, \lambda + 2, 2\lambda + 2)$.
Let $A = (1, 2, 12)$. The vector $\vec{AF} = (2\lambda + 1, \lambda, 2\lambda - 10)$.
Since $AF \perp PQ$,$\vec{AF} \cdot (2, 1, 2) = 0$.
$2(2\lambda + 1) + 1(\lambda) + 2(2\lambda - 10) = 0$.
$4\lambda + 2 + \lambda + 4\lambda - 20 = 0 \Rightarrow 9\lambda = 18 \Rightarrow \lambda = 2$.
So $F = (6, 4, 6)$.
The length $l = AF = \sqrt{(6 - 1)^2 + (4 - 2)^2 + (6 - 12)^2} = \sqrt{5^2 + 2^2 + (-6)^2} = \sqrt{25 + 4 + 36} = \sqrt{65}$.
Therefore,$l^2 = 65$.

(B) Let the lines be $L_1: \frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}=\lambda$ and $L_2: \frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}=\mu$.
Any point $M$ on $L_1$ is $(4\lambda+5, \lambda+4, 3\lambda+5)$ and any point $N$ on $L_2$ is $(12\mu-8, 5\mu-2, 9\mu-11)$.
The vector $\overrightarrow{MN} = (12\mu-4\lambda-13, 5\mu-\lambda-6, 9\mu-3\lambda-16)$.
The direction vectors are $\vec{b}_1 = (4, 1, 3)$ and $\vec{b}_2 = (12, 5, 9)$.
The shortest distance vector $\overrightarrow{MN}$ must be perpendicular to both $\vec{b}_1$ and $\vec{b}_2$.
$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 3 \\ 12 & 5 & 9 \end{vmatrix} = -6\hat{i} + 0\hat{j} + 8\hat{k}$.
Since $\overrightarrow{MN}$ is parallel to $\vec{b}_1 \times \vec{b}_2$,we have $\frac{12\mu-4\lambda-13}{-6} = \frac{5\mu-\lambda-6}{0} = \frac{9\mu-3\lambda-16}{8}$.
From the middle term,$5\mu-\lambda-6=0 \implies \lambda = 5\mu-6$.
Substituting into the ratio $\frac{12\mu-4\lambda-13}{-6} = \frac{9\mu-3\lambda-16}{8}$:
$8(12\mu-4(5\mu-6)-13) = -6(9\mu-3(5\mu-6)-16) \implies 8(-8\mu+11) = -6(-6\mu+2) \implies -64\mu+88 = 36\mu-12 \implies 100\mu = 100 \implies \mu=1$.
Then $\lambda = 5(1)-6 = -1$.
Thus,$M = (4(-1)+5, -1+4, 3(-1)+5) = (1, 3, 2)$ and $N = (12(1)-8, 5(1)-2, 9(1)-11) = (4, 3, -2)$.
Finally,$\overrightarrow{OM} \cdot \overrightarrow{ON} = (1)(4) + (3)(3) + (2)(-2) = 4 + 9 - 4 = 9$.

(B) Let the line be $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = k$.
Any point $P$ on the line is given by $P(5k-3, 2k+1, 3k-4)$.
The direction ratios of the line $AP$,where $A$ is $(1, 2, 3)$,are $(5k-3-1, 2k+1-2, 3k-4-3) = (5k-4, 2k-1, 3k-7)$.
The direction ratios of the given line are $(5, 2, 3)$.
Since $AP$ is perpendicular to the line,the dot product of their direction ratios must be zero:
$5(5k-4) + 2(2k-1) + 3(3k-7) = 0$
$25k - 20 + 4k - 2 + 9k - 21 = 0$
$38k - 43 = 0 \implies k = \frac{43}{38}$.
The coordinates of the foot $P$ are $(\alpha, \beta, \gamma) = (5k-3, 2k+1, 3k-4)$.
Then $\alpha + \beta + \gamma = (5k-3) + (2k+1) + (3k-4) = 10k - 6$.
Substituting $k = \frac{43}{38}$:
$\alpha + \beta + \gamma = 10\left(\frac{43}{38}\right) - 6 = \frac{430 - 228}{38} = \frac{202}{38} = \frac{101}{19}$.
Therefore,$19(\alpha + \beta + \gamma) = 19 \times \frac{101}{19} = 101$.

(B) For the first pair of lines:
$L_1: \frac{x+1}{1} = \frac{y}{1/2} = \frac{z}{-1/12} \implies \vec{a}_1 = (-1, 0, 0), \vec{b}_1 = (1, 1/2, -1/12)$
$L_2: \frac{x}{1} = \frac{y+2}{1} = \frac{z-1}{1/6} \implies \vec{a}_2 = (0, -2, 1), \vec{b}_2 = (1, 1, 1/6)$
Shortest distance $d_1 = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} = 2$.
For the second pair of lines:
$L_3: \frac{x-1}{2} = \frac{y+8}{-7} = \frac{z-4}{5}, \vec{a}_3 = (1, -8, 4), \vec{b}_3 = (2, -7, 5)$
$L_4: \frac{x-1}{2} = \frac{y-2}{1} = \frac{z-6}{-3}, \vec{a}_4 = (1, 2, 6), \vec{b}_4 = (2, 1, -3)$
Shortest distance $d_2 = \frac{|(\vec{a}_4 - \vec{a}_3) \cdot (\vec{b}_3 \times \vec{b}_4)|}{|\vec{b}_3 \times \vec{b}_4|} = \frac{12}{\sqrt{3}}$.
Calculating the final value:
$\frac{32 \sqrt{3} d_1}{d_2} = \frac{32 \sqrt{3} \times 2}{12/\sqrt{3}} = \frac{64 \sqrt{3} \times \sqrt{3}}{12} = \frac{64 \times 3}{12} = \frac{192}{12} = 16$.

(B) Let the point $M$ on $L_1$ be $(3\lambda+1, 2\lambda+2, -2\lambda-1)$.
Then $\alpha+\beta+\gamma = (3\lambda+1) + (2\lambda+2) + (-2\lambda-1) = 3\lambda+2$.
Let the point $N$ on $L_2$ be $(-3\mu-2, -2\mu+2, 4\mu+1)$.
Then $a+b+c = (-3\mu-2) + (-2\mu+2) + (4\mu+1) = -\mu+1$.
The line passes through $P(-1, 2, 3)$,$M$,and $N$. Thus,the vectors $\vec{PM}$ and $\vec{PN}$ are collinear.
$\vec{PM} = (3\lambda+2, 2\lambda, -2\lambda-4)$ and $\vec{PN} = (-3\mu-1, -2\mu, 4\mu-2)$.
Since they are collinear,$\frac{3\lambda+2}{-3\mu-1} = \frac{2\lambda}{-2\mu} = \frac{-2\lambda-4}{4\mu-2}$.
From $\frac{2\lambda}{-2\mu} = \frac{3\lambda+2}{-3\mu-1}$,we get $\lambda(-3\mu-1) = -\mu(3\lambda+2) \Rightarrow -3\lambda\mu - \lambda = -3\lambda\mu - 2\mu \Rightarrow \lambda = 2\mu$.
From $\frac{2\lambda}{-2\mu} = \frac{-2\lambda-4}{4\mu-2}$,we get $\frac{\lambda}{-\mu} = \frac{-\lambda-2}{2\mu-1} \Rightarrow 2\lambda\mu - \lambda = \lambda\mu + 2\mu \Rightarrow \lambda\mu = \lambda + 2\mu$.
Substituting $\lambda = 2\mu$,we get $(2\mu)\mu = 2\mu + 2\mu \Rightarrow 2\mu^2 = 4\mu \Rightarrow \mu = 2$ (as $\mu \neq 0$).
Then $\lambda = 4$.
Thus,$\alpha+\beta+\gamma = 3(4)+2 = 14$ and $a+b+c = -(2)+1 = -1$.
Therefore,$\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2} = \frac{14^2}{(-1)^2} = 196$.

(D) The direction vector of the required line is the cross product of the direction vectors of the two given lines:
$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix} = \hat{i}(6 - 15) - \hat{j}(4 + 5) + \hat{k}(6 + 3) = -9\hat{i} - 9\hat{j} + 9\hat{k}$.
We can take the direction vector as $\vec{d} = \hat{i} + \hat{j} - \hat{k}$.
The equation of the line passing through $P(5, -4, 3)$ is $\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} - \hat{k})$.
Any point $M$ on this line is $(5+\lambda, -4+\lambda, 3-\lambda)$.
Let $Q$ be $(0, 2, -2)$. The vector $\vec{QM} = (5+\lambda - 0)\hat{i} + (-4+\lambda - 2)\hat{j} + (3-\lambda + 2)\hat{k} = (5+\lambda)\hat{i} + (\lambda-6)\hat{j} + (5-\lambda)\hat{k}$.
Since $QM \perp$ line,$\vec{QM} \cdot (\hat{i} + \hat{j} - \hat{k}) = 0$.
$(5+\lambda)(1) + (\lambda-6)(1) + (5-\lambda)(-1) = 0 \implies 5+\lambda + \lambda-6 - 5+\lambda = 0 \implies 3\lambda - 6 = 0 \implies \lambda = 2$.
The point $M$ is $(5+2, -4+2, 3-2) = (7, -2, 1)$.
The distance $QM = \sqrt{(7-0)^2 + (-2-2)^2 + (1 - (-2))^2} = \sqrt{49 + 16 + 9} = \sqrt{74}$.

(B) Let $P = (2, 3, 5)$ and $R = (\alpha, \beta, \gamma)$ be its mirror image in the line $L: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$.
Let $M$ be the midpoint of $PR$. Since $R$ is the mirror image of $P$ in the line $L$,the line segment $PR$ is perpendicular to the line $L$.
The direction vector of the line $L$ is $\vec{v} = (2, 3, 4)$.
The vector $\vec{PR} = (\alpha - 2, \beta - 3, \gamma - 5)$.
Since $\vec{PR} \perp \vec{v}$,their dot product must be zero:
$\vec{PR} \cdot \vec{v} = 0$
$(\alpha - 2, \beta - 3, \gamma - 5) \cdot (2, 3, 4) = 0$
$2(\alpha - 2) + 3(\beta - 3) + 4(\gamma - 5) = 0$
$2\alpha - 4 + 3\beta - 9 + 4\gamma - 20 = 0$
$2\alpha + 3\beta + 4\gamma = 4 + 9 + 20$
$2\alpha + 3\beta + 4\gamma = 33$.

(C) First,find the equation of line $L_2$. The direction vector of $L_2$ is perpendicular to both the vector $\vec{AB} = (-1 - (-4), 6 - 4, 3 - 3) = (3, 2, 0)$ and the direction vector of the given line $\vec{v} = (-2, 3, 1)$.
Thus,the direction vector $\vec{n_2}$ of $L_2$ is $\vec{AB} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 0 \\ -2 & 3 & 1 \end{vmatrix} = \hat{i}(2-0) - \hat{j}(3-0) + \hat{k}(9-(-4)) = 2\hat{i} - 3\hat{j} + 13\hat{k}$.
However,the problem states $L_2$ passes through $A(-4, 4, 3)$ and $B(-1, 6, 3)$,so its direction vector is simply $\vec{AB} = (3, 2, 0)$.
The shortest distance $d$ between $L_1: \vec{r} = (1, -1, -4) + \lambda(2, -3, 2)$ and $L_2: \vec{r} = (-4, 4, 3) + \mu(3, 2, 0)$ is given by $d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{n_1} \times \vec{n_2})|}{ |\vec{n_1} \times \vec{n_2}| }$.
$\vec{a_2} - \vec{a_1} = (-4-1, 4-(-1), 3-(-4)) = (-5, 5, 7)$.
$\vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{vmatrix} = \hat{i}(0-4) - \hat{j}(0-6) + \hat{k}(4-(-9)) = -4\hat{i} + 6\hat{j} + 13\hat{k}$.
$|\vec{n_1} \times \vec{n_2}| = \sqrt{(-4)^2 + 6^2 + 13^2} = \sqrt{16 + 36 + 169} = \sqrt{221}$.
$|(\vec{a_2} - \vec{a_1}) \cdot (\vec{n_1} \times \vec{n_2})| = |(-5)(-4) + (5)(6) + (7)(13)| = |20 + 30 + 91| = 141$.
Therefore,$d = \frac{141}{\sqrt{221}}$.

(D) The direction vector of line $AB$ is $\vec{v} = (16-4, -2-(-6), 4-(-2)) = (12, 4, 6)$.
The unit vector in the direction of $AB$ is $\hat{u} = \frac{(12, 4, 6)}{\sqrt{12^2 + 4^2 + 6^2}} = \frac{(12, 4, 6)}{\sqrt{144 + 16 + 36}} = \frac{(12, 4, 6)}{\sqrt{196}} = \frac{(12, 4, 6)}{14} = (\frac{6}{7}, \frac{2}{7}, \frac{3}{7})$.
Point $P$ is at a distance of $21$ units from $A(4, -6, -2)$ along the line $AB$,so $P = A + 21 \hat{u}$.
$P = (4, -6, -2) + 21(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}) = (4 + 18, -6 + 6, -2 + 9) = (22, 0, 7)$.
Here $a=22, b=0, c=7$,which are non-negative integers.
The distance between $P(22, 0, 7)$ and $Q(4, -12, 3)$ is $\sqrt{(22-4)^2 + (0-(-12))^2 + (7-3)^2}$.
$= \sqrt{18^2 + 12^2 + 4^2} = \sqrt{324 + 144 + 16} = \sqrt{484} = 22$.

(B) The lines are $L_1: \frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $L_2: \frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$.
Passing points are $A = (\lambda, 2, 1)$ and $B = (\sqrt{3}, 1, 2)$.
Direction vectors are $\vec{v_1} = -2\hat{i} + \hat{j} + \hat{k}$ and $\vec{v_2} = \hat{i} - 2\hat{j} + \hat{k}$.
The cross product $\vec{n} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(1+2) - \hat{j}(-2-1) + \hat{k}(4-1) = 3\hat{i} + 3\hat{j} + 3\hat{k}$.
The magnitude is $|\vec{n}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$.
The shortest distance $d$ is given by $d = \frac{|(\vec{B}-\vec{A}) \cdot \vec{n}|}{|\vec{n}|}$.
$\vec{B}-\vec{A} = (\sqrt{3}-\lambda)\hat{i} - \hat{j} + \hat{k}$.
$d = \frac{|3(\sqrt{3}-\lambda) - 3 + 3|}{3\sqrt{3}} = \frac{|3(\sqrt{3}-\lambda)|}{3\sqrt{3}} = \frac{|\sqrt{3}-\lambda|}{\sqrt{3}}$.
Given $d = 1$,so $|\sqrt{3}-\lambda| = \sqrt{3}$.
This implies $\sqrt{3}-\lambda = \sqrt{3}$ or $\sqrt{3}-\lambda = -\sqrt{3}$.
Thus,$\lambda = 0$ or $\lambda = 2\sqrt{3}$.
The sum of all possible values of $\lambda$ is $0 + 2\sqrt{3} = 2\sqrt{3}$.

(A) Let $P$ be a point on $L_1$ given by $(1+\lambda, 2-\lambda, 3+\lambda)$ and $Q$ be a point on $L_2$ given by $(4+\mu, 5+\mu, 6-\mu)$.
The vector $\vec{PQ} = (4+\mu-(1+\lambda))\hat{i} + (5+\mu-(2-\lambda))\hat{j} + (6-\mu-(3+\lambda))\hat{k} = (3+\mu-\lambda)\hat{i} + (3+\mu+\lambda)\hat{j} + (3-\mu-\lambda)\hat{k}$.
The direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = \hat{i}-\hat{j}+\hat{k}$ and $\vec{v_2} = \hat{i}+\hat{j}-\hat{k}$ respectively.
Since $PQ$ is the shortest distance line,it is perpendicular to both $L_1$ and $L_2$. Thus,$\vec{PQ} \cdot \vec{v_1} = 0$ and $\vec{PQ} \cdot \vec{v_2} = 0$.
$\vec{PQ} \cdot \vec{v_1} = (3+\mu-\lambda) - (3+\mu+\lambda) + (3-\mu-\lambda) = 3 - 3\lambda - \mu = 0 \implies 3\lambda + \mu = 3$.
$\vec{PQ} \cdot \vec{v_2} = (3+\mu-\lambda) + (3+\mu+\lambda) - (3-\mu-\lambda) = 3 + 3\mu + \lambda = 0 \implies \lambda + 3\mu = -3$.
Solving these equations: $3(3\lambda + \mu) - (\lambda + 3\mu) = 3(3) - (-3) \implies 8\lambda = 12 \implies \lambda = \frac{3}{2}$.
Substituting $\lambda = \frac{3}{2}$ into $3\lambda + \mu = 3$,we get $\frac{9}{2} + \mu = 3 \implies \mu = -\frac{3}{2}$.
Coordinates of $P = (1+\frac{3}{2}, 2-\frac{3}{2}, 3+\frac{3}{2}) = (\frac{5}{2}, \frac{1}{2}, \frac{9}{2})$.
Coordinates of $Q = (4-\frac{3}{2}, 5-\frac{3}{2}, 6+\frac{3}{2}) = (\frac{5}{2}, \frac{7}{2}, \frac{15}{2})$.
The midpoint $(\alpha, \beta, \gamma) = (\frac{5/2+5/2}{2}, \frac{1/2+7/2}{2}, \frac{9/2+15/2}{2}) = (\frac{5}{2}, 2, 6)$.
Thus,$2(\alpha+\beta+\gamma) = 2(\frac{5}{2} + 2 + 6) = 5 + 4 + 12 = 21$.

(C) The general point on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2} = \lambda$ is given by $P, Q = (8\lambda-3, 2\lambda+4, 2\lambda-1)$.
The distance from this point to $R(1,2,3)$ is $6$ units,so the square of the distance is $36$:
$(8\lambda-3-1)^2 + (2\lambda+4-2)^2 + (2\lambda-1-3)^2 = 36$
$(8\lambda-4)^2 + (2\lambda+2)^2 + (2\lambda-4)^2 = 36$
$64(\lambda^2 - \lambda + \frac{1}{4}) + 4(\lambda^2 + 2\lambda + 1) + 4(\lambda^2 - 4\lambda + 4) = 36$
$64\lambda^2 - 64\lambda + 16 + 4\lambda^2 + 8\lambda + 4 + 4\lambda^2 - 16\lambda + 16 = 36$
$72\lambda^2 - 72\lambda + 36 = 36$
$72\lambda(\lambda - 1) = 0$
Thus,$\lambda = 0$ or $\lambda = 1$.
For $\lambda = 0$,the point is $P(-3, 4, -1)$.
For $\lambda = 1$,the point is $Q(5, 6, 1)$.
The centroid of $\Delta PQR$ is $(\frac{-3+5+1}{3}, \frac{4+6+2}{3}, \frac{-1+1+3}{3}) = (1, 4, 1) = (\alpha, \beta, \gamma)$.
Therefore,$\alpha^2 + \beta^2 + \gamma^2 = 1^2 + 4^2 + 1^2 = 1 + 16 + 1 = 18$.

(C) Let the line be $L: \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1} = \lambda$. Any point $N$ on the line is $(3\lambda+1, 2\lambda-1, \lambda+2)$.
Since $PN$ is perpendicular to the line with direction vector $\vec{b} = (3, 2, 1)$,the vector $\vec{PN} = (3\lambda+1-3, 2\lambda-1-4, \lambda+2-9) = (3\lambda-2, 2\lambda-5, \lambda-7)$.
Since $\vec{PN} \cdot \vec{b} = 0$,we have $3(3\lambda-2) + 2(2\lambda-5) + 1(\lambda-7) = 0$.
$9\lambda - 6 + 4\lambda - 10 + \lambda - 7 = 0 \Rightarrow 14\lambda = 23 \Rightarrow \lambda = \frac{23}{14}$.
Substituting $\lambda$ into the coordinates of $N$,we get $N = \left(3(\frac{23}{14})+1, 2(\frac{23}{14})-1, \frac{23}{14}+2\right) = \left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}\right)$.
Let the mirror image be $A(\alpha, \beta, \gamma)$. Since $N$ is the midpoint of $PA$,$\frac{\alpha+3}{2} = \frac{83}{14} \Rightarrow \alpha = \frac{83}{7} - 3 = \frac{62}{7}$.
$\frac{\beta+4}{2} = \frac{32}{14} \Rightarrow \beta = \frac{32}{7} - 4 = \frac{4}{7}$.
$\frac{\gamma+9}{2} = \frac{51}{14} \Rightarrow \gamma = \frac{51}{7} - 9 = \frac{-12}{7}$.
Thus,$14(\alpha+\beta+\gamma) = 14(\frac{62+4-12}{7}) = 2(54) = 108$.

(A) The direction vector of the line passing through $P(1, -2, 3)$ and $Q(5, -4, 7)$ is $\vec{v} = (5-1, -4-(-2), 7-3) = (4, -2, 4)$.
The equation of the line is $\frac{x-1}{4} = \frac{y+2}{-2} = \frac{z-3}{4} = t$.
Any point on this line is given by $(4t+1, -2t-2, 4t+3)$.
The distance of this point from $P(1, -2, 3)$ is $\sqrt{(4t+1-1)^2 + (-2t-2+2)^2 + (4t+3-3)^2} = \sqrt{16t^2 + 4t^2 + 16t^2} = \sqrt{36t^2} = 6|t|$.
Given the distance is $9$ units,$6|t| = 9$,so $t = \pm \frac{3}{2}$.
For $t = \frac{3}{2}$,the point is $(4(\frac{3}{2})+1, -2(\frac{3}{2})-2, 4(\frac{3}{2})+3) = (7, -5, 9)$.
For $t = -\frac{3}{2}$,the point is $(4(-\frac{3}{2})+1, -2(-\frac{3}{2})-2, 4(-\frac{3}{2})+3) = (-5, 1, -3)$.
The distance of $(7, -5, 9)$ from the origin is $\sqrt{7^2 + (-5)^2 + 9^2} = \sqrt{49 + 25 + 81} = \sqrt{155}$.
The distance of $(-5, 1, -3)$ from the origin is $\sqrt{(-5)^2 + 1^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35}$.
Since the point is farther from the origin,we choose $(7, -5, 9)$.
Thus,$\alpha^2 + \beta^2 + \gamma^2 = 7^2 + (-5)^2 + 9^2 = 49 + 25 + 81 = 155$.

(D) The shortest distance 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 $d = \frac{|(\vec{r_2}-\vec{r_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$.
Here,$\vec{r_1} = (-2, -3, 5)$,$\vec{r_2} = (3, 2, -4)$,$\vec{b_1} = (2, 3, 4)$,and $\vec{b_2} = (1, -3, 2)$.
$\vec{r_2}-\vec{r_1} = (5, 5, -9)$.
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & -3 & 2 \end{vmatrix} = \hat{i}(6+12) - \hat{j}(4-4) + \hat{k}(-6-3) = 18\hat{i} - 9\hat{k}$.
$|\vec{b_1} \times \vec{b_2}| = \sqrt{18^2 + (-9)^2} = \sqrt{324 + 81} = \sqrt{405} = 9\sqrt{5}$.
Shortest distance $d = \frac{|(5, 5, -9) \cdot (18, 0, -9)|}{9\sqrt{5}} = \frac{|90 + 0 + 81|}{9\sqrt{5}} = \frac{171}{9\sqrt{5}} = \frac{19}{\sqrt{5}}$.
Given $d = \frac{38}{3\sqrt{5}}k$,so $\frac{19}{\sqrt{5}} = \frac{38}{3\sqrt{5}}k \Rightarrow k = \frac{19 \times 3}{38} = \frac{3}{2}$.
Now,$\int_0^{3/2} [x^2] dx = \int_0^1 [x^2] dx + \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{3/2} [x^2] dx = 0 + \int_1^{\sqrt{2}} 1 dx + \int_{\sqrt{2}}^{3/2} 2 dx = (\sqrt{2}-1) + 2(\frac{3}{2}-\sqrt{2}) = \sqrt{2}-1+3-2\sqrt{2} = 2-\sqrt{2}$.
Comparing with $\alpha-\sqrt{\alpha}$,we get $\alpha=2$.
Thus,$6\alpha^3 = 6(2^3) = 6 \times 8 = 48$.

(C) Let the first line be $L_1: \frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}=\lambda$. Any point on $L_1$ is $P(\lambda+2, 5\lambda+4, \lambda+2)$.
Let the second line be $L_2: \frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}=\mu$. Any point on $L_2$ is $P(2\mu+3, 3\mu+2, 2\mu+3)$.
For intersection,$\lambda+2 = 2\mu+3$ and $5\lambda+4 = 3\mu+2$.
From the first equation,$\lambda = 2\mu+1$. Substituting into the second: $5(2\mu+1)+4 = 3\mu+2 \implies 10\mu+9 = 3\mu+2 \implies 7\mu = -7 \implies \mu = -1$.
Then $\lambda = 2(-1)+1 = -1$. The point of intersection $P$ is $(1, -1, 1)$.
The line $L_3$ is $4x=2y=z$,which can be written as $\frac{x}{1/4} = \frac{y}{1/2} = \frac{z}{1}$ or $\frac{x}{1} = \frac{y}{2} = \frac{z}{4} = k$.
Any point $Q$ on $L_3$ is $(k, 2k, 4k)$. The vector $\vec{PQ} = (k-1, 2k+1, 4k-1)$.
Since $\vec{PQ}$ is perpendicular to the direction vector of $L_3$,which is $\vec{v} = (1, 2, 4)$,we have $\vec{PQ} \cdot \vec{v} = 0$.
$(k-1)(1) + (2k+1)(2) + (4k-1)(4) = 0 \implies k-1 + 4k+2 + 16k-4 = 0 \implies 21k - 3 = 0 \implies k = \frac{1}{7}$.
The point $Q$ is $(\frac{1}{7}, \frac{2}{7}, \frac{4}{7})$.
The distance $PQ = \sqrt{(1-\frac{1}{7})^2 + (-1-\frac{2}{7})^2 + (1-\frac{4}{7})^2} = \sqrt{(\frac{6}{7})^2 + (-\frac{9}{7})^2 + (\frac{3}{7})^2} = \sqrt{\frac{36+81+9}{49}} = \sqrt{\frac{126}{49}} = \frac{\sqrt{9 \times 14}}{7} = \frac{3\sqrt{14}}{7}$.