Line Questions in English

(B) Two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ are perpendicular if $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
Given the lines are $\frac{x-5}{7}=\frac{y-5}{k}=\frac{z-2}{1}$ and $\frac{x}{1}=\frac{y-3}{2}=\frac{z+1}{3}$.
The direction ratios of the first line are $(7, k, 1)$ and the direction ratios of the second line are $(1, 2, 3)$.
Since the lines are perpendicular,we have:
$(7)(1) + (k)(2) + (1)(3) = 0$
$7 + 2k + 3 = 0$
$10 + 2k = 0$
$2k = -10$
$k = -5$.
Thus,the correct option is $B$.

(D) The equation of a line passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.
Substituting the given points $(2, -3, 1)$ and $(3, -4, -5)$:
$\frac{x-2}{3-2} = \frac{y-(-3)}{-4-(-3)} = \frac{z-1}{-5-1}$
$\frac{x-2}{1} = \frac{y+3}{-1} = \frac{z-1}{-6} = k$ (let).
For the point $(0, a, b)$ to lie on the line,we set $x = 0$:
$\frac{0-2}{1} = k \implies k = -2$.
Now,find $a$ and $b$ using $k = -2$:
$\frac{a+3}{-1} = -2 \implies a+3 = 2 \implies a = -1$.
$\frac{b-1}{-6} = -2 \implies b-1 = 12 \implies b = 13$.
Therefore,$a+b = -1 + 13 = 12$.

(C) First,rewrite the equations of the lines in standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Line $1$: $\frac{x-2}{2} = \frac{y-2}{-3} = \frac{z-1}{2}$. The direction vector is $\vec{v_1} = 2\hat{i} - 3\hat{j} + 2\hat{k}$.
Line $2$: $\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-3}$. The direction vector is $\vec{v_2} = 2\hat{i} + 1\hat{j} - 3\hat{k}$.
The cosine of the angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{v_1} \cdot \vec{v_2}|}{|\vec{v_1}| |\vec{v_2}|}$.
$\vec{v_1} \cdot \vec{v_2} = (2)(2) + (-3)(1) + (2)(-3) = 4 - 3 - 6 = -5$.
$|\vec{v_1}| = \sqrt{2^2 + (-3)^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17}$.
$|\vec{v_2}| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$.
$\cos \theta = \frac{|-5|}{\sqrt{17} \sqrt{14}} = \frac{5}{\sqrt{238}}$.
Using $\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{25}{238} = \frac{213}{238}$.
Thus,$\sin \theta = \sqrt{\frac{213}{238}}$,which implies $\theta = \sin^{-1}\left(\sqrt{\frac{213}{238}}\right)$.

(A) The given lines are $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-5}{1}=\frac{z-6}{-5}$.
The direction ratios of the first line are $\vec{b_1} = -3\hat{i} + 2k\hat{j} + 2\hat{k}$.
The direction ratios of the second line are $\vec{b_2} = 3k\hat{i} + 1\hat{j} - 5\hat{k}$.
Since the lines are mutually perpendicular,their dot product must be zero,i.e.,$\vec{b_1} \cdot \vec{b_2} = 0$.
$(-3)(3k) + (2k)(1) + (2)(-5) = 0$.
$-9k + 2k - 10 = 0$.
$-7k - 10 = 0$.
$-7k = 10$.
$k = -\frac{10}{7}$.
Thus,the value of $k$ is $-\frac{10}{7}$.

(B) Let the required line pass through the point $P(0,1,2)$ with direction ratios $(a, b, c)$.
The equation of the line is $\frac{x-0}{a} = \frac{y-1}{b} = \frac{z-2}{c}$.
The given line is $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{-2}$,which has direction ratios $(2, 3, -2)$.
Since the two lines are perpendicular,the dot product of their direction ratios must be zero:
$2a + 3b - 2c = 0$.
Checking the options:
For option $C$: The direction ratios are $(3, 4, 3)$.
$2(3) + 3(4) - 2(3) = 6 + 12 - 6 = 12 \neq 0$.
For option $B$: The direction ratios are $(-3, 4, 3)$.
$2(-3) + 3(4) - 2(3) = -6 + 12 - 6 = 0$.
Thus,the line $\frac{x}{-3} = \frac{y-1}{4} = \frac{z-2}{3}$ is perpendicular to the given line.

(C) Let the foot of the perpendicular drawn from the point $P(3, -1, 11)$ to the line be $L$.
Since $L$ lies on the line $\frac{x}{2} = \frac{y-2}{3} = \frac{z-3}{4} = t$,the coordinates of $L$ are $(2t, 3t+2, 4t+3)$.
The direction ratios of the line $PL$ are $(2t-3, 3t+2-(-1), 4t+3-11)$,which simplifies to $(2t-3, 3t+3, 4t-8)$.
Since $PL$ is perpendicular to the line with direction ratios $(2, 3, 4)$,their dot product must be zero:
$2(2t-3) + 3(3t+3) + 4(4t-8) = 0$.
$4t - 6 + 9t + 9 + 16t - 32 = 0$.
$29t - 29 = 0 \implies t = 1$.
Substituting $t=1$ into the coordinates of $L$,we get $L(2, 5, 7)$.
The length of the perpendicular $PL$ is the distance between $P(3, -1, 11)$ and $L(2, 5, 7)$:
$PL = \sqrt{(2-3)^2 + (5-(-1))^2 + (7-11)^2} = \sqrt{(-1)^2 + 6^2 + (-4)^2} = \sqrt{1 + 36 + 16} = \sqrt{53}$.

(D) The direction ratios of the first line are $\vec{b_1} = \langle 3, 5, 4 \rangle$.
The direction ratios of the second line are $\vec{b_2} = \langle 1, 4, 2 \rangle$.
The angle $\theta$ between two lines with direction ratios $\langle a_1, b_1, c_1 \rangle$ and $\langle a_2, b_2, c_2 \rangle$ is given by $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$.
Substituting the values:
$\cos \theta = \left| \frac{(3)(1) + (5)(4) + (4)(2)}{\sqrt{3^2 + 5^2 + 4^2} \sqrt{1^2 + 4^2 + 2^2}} \right|$
$\cos \theta = \left| \frac{3 + 20 + 8}{\sqrt{9 + 25 + 16} \sqrt{1 + 16 + 4}} \right|$
$\cos \theta = \left| \frac{31}{\sqrt{50} \sqrt{21}} \right| = \frac{31}{5 \sqrt{2} \sqrt{21}} = \frac{31}{5 \sqrt{42}}$.
Thus,$\theta = \cos^{-1} \left( \frac{31}{5 \sqrt{42}} \right)$.
Since this value is not present in the given options,the correct choice is $D$.

(B) Given points are $A(-3, 4, 11)$ and $B(1, -2, 7)$.
Direction ratios of the line $AB$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (1 - (-3), -2 - 4, 7 - 11) = (4, -6, -4)$.
Dividing by $-2$,we get the simplified direction ratios as $(-2, 3, 2)$.
The 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 point $(-3, 4, 11)$ and direction ratios $(-2, 3, 2)$,we get $\frac{x - (-3)}{-2} = \frac{y - 4}{3} = \frac{z - 11}{2}$.
Thus,the equation is $\frac{x + 3}{-2} = \frac{y - 4}{3} = \frac{z - 11}{2}$.

(A) Let the given line be $\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-3}{2} = \lambda$.
Any point $A$ on the line is given by $A = (2\lambda + 1, \lambda + 2, 2\lambda + 3)$.
The vector $\vec{PA} = (2\lambda + 1 - 1)\hat{i} + (\lambda + 2 - 2)\hat{j} + (2\lambda + 3 - 1)\hat{k} = 2\lambda\hat{i} + \lambda\hat{j} + (2\lambda + 2)\hat{k}$.
The direction vector of the line is $\vec{v} = 2\hat{i} + 1\hat{j} + 2\hat{k}$.
Since $\vec{PA}$ is perpendicular to the line,$\vec{PA} \cdot \vec{v} = 0$.
$(2\lambda)(2) + (\lambda)(1) + (2\lambda + 2)(2) = 0$.
$4\lambda + \lambda + 4\lambda + 4 = 0 \implies 9\lambda = -4 \implies \lambda = -\frac{4}{9}$.
Substituting $\lambda = -\frac{4}{9}$ in $A$,we get $A = (2(-\frac{4}{9}) + 1, -\frac{4}{9} + 2, 2(-\frac{4}{9}) + 3) = (\frac{1}{9}, \frac{14}{9}, \frac{19}{9})$.
The distance $PA = \sqrt{(\frac{1}{9} - 1)^2 + (\frac{14}{9} - 2)^2 + (\frac{19}{9} - 1)^2} = \sqrt{(-\frac{8}{9})^2 + (-\frac{4}{9})^2 + (\frac{10}{9})^2}$.
$PA = \sqrt{\frac{64 + 16 + 100}{81}} = \sqrt{\frac{180}{81}} = \sqrt{\frac{20}{9}} = \frac{2\sqrt{5}}{3}$.

(A) Let the given line be $ \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3} = \lambda $.
Any point $ R $ on the line is given by $ R(\lambda, 1+2\lambda, 2+3\lambda) $.
Let $ P $ be the point $ (1,6,3) $. The vector $ \vec{PR} $ is $ (\lambda-1, 2\lambda-5, 3\lambda-1) $.
Since $ PR $ is perpendicular to the line with direction ratios $ (1,2,3) $,their dot product is 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 $ in $ R $,we get $ R(1, 3, 5) $.
Since $ R $ is the midpoint of $ PQ $,where $ Q(x_1, y_1, z_1) $ is the image of $ P $,
$ \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 $.
Thus,the image $ Q $ is $ (1,0,7) $.

(C) The given lines are $2x = 3y = -z$ and $6x = -y = -4z$.
First,we write the equations in the standard form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
For the first line $2x = 3y = -z$,dividing by $6$,we get $\frac{x}{3} = \frac{y}{2} = \frac{z}{-6}$. Thus,the direction ratios are $\vec{v_1} = (3, 2, -6)$.
For the second line $6x = -y = -4z$,dividing by $12$,we get $\frac{x}{2} = \frac{y}{-12} = \frac{z}{-3}$. Thus,the direction ratios are $\vec{v_2} = (2, -12, -3)$.
Two lines are perpendicular if the dot product of their direction vectors is zero,i.e.,$a_1a_2 + b_1b_2 + c_1c_2 = 0$.
Calculating the dot product: $(3)(2) + (2)(-12) + (-6)(-3) = 6 - 24 + 18 = 0$.
Since the dot product is $0$,the lines are perpendicular to each other.
Therefore,the angle between the lines is $90^{\circ}$.

(B) Let the given point be $A(1, 2, -4)$ and the line be $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6} = \lambda$.
Any point $P$ on the line is given by $(2\lambda+3, 3\lambda+3, 6\lambda-5)$.
The direction ratios of the line $AP$ are $(2\lambda+3-1, 3\lambda+3-2, 6\lambda-5-(-4))$,which simplifies to $(2\lambda+2, 3\lambda+1, 6\lambda-1)$.
Since $AP$ is perpendicular to the given line with direction ratios $(2, 3, 6)$,their dot product must be zero:
$2(2\lambda+2) + 3(3\lambda+1) + 6(6\lambda-1) = 0$
$4\lambda + 4 + 9\lambda + 3 + 36\lambda - 6 = 0$
$49\lambda + 1 = 0 \Rightarrow \lambda = -\frac{1}{49}$.
The square of the distance $AP$ is $AP^2 = (2\lambda+2)^2 + (3\lambda+1)^2 + (6\lambda-1)^2$.
Substituting $\lambda = -\frac{1}{49}$:
$AP^2 = 49\lambda^2 + 2\lambda + 6 = 49(-\frac{1}{49})^2 + 2(-\frac{1}{49}) + 6$
$AP^2 = \frac{1}{49} - \frac{2}{49} + 6 = 6 - \frac{1}{49} = \frac{294-1}{49} = \frac{293}{49}$.
Therefore,$AP = \sqrt{\frac{293}{49}} = \frac{\sqrt{293}}{7}$.

(B) Given line: $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} = \lambda$ $(1)$
Point $P(-2, 4, -5)$.
Any point $Q$ on the line is given by $Q(3\lambda - 3, 5\lambda + 4, 6\lambda - 8)$.
The direction ratios of the line $PQ$ are $(3\lambda - 3 - (-2), 5\lambda + 4 - 4, 6\lambda - 8 - (-5)) = (3\lambda - 1, 5\lambda, 6\lambda - 3)$.
Since $PQ$ is perpendicular to the given line with direction ratios $(3, 5, 6)$,their dot product is zero:
$3(3\lambda - 1) + 5(5\lambda) + 6(6\lambda - 3) = 0$
$9\lambda - 3 + 25\lambda + 36\lambda - 18 = 0$
$70\lambda - 21 = 0 \Rightarrow \lambda = \frac{21}{70} = \frac{3}{10}$.
Substituting $\lambda = \frac{3}{10}$ in $Q$,we get $Q\left(3(\frac{3}{10}) - 3, 5(\frac{3}{10}) + 4, 6(\frac{3}{10}) - 8\right) = Q\left(-\frac{21}{10}, \frac{55}{10}, -\frac{62}{10}\right)$.
The distance $PQ = \sqrt{(-\frac{21}{10} + 2)^2 + (\frac{55}{10} - 4)^2 + (-\frac{62}{10} + 5)^2}$
$PQ = \sqrt{(-\frac{1}{10})^2 + (\frac{15}{10})^2 + (-\frac{12}{10})^2} = \sqrt{\frac{1 + 225 + 144}{100}} = \sqrt{\frac{370}{100}} = \sqrt{\frac{37}{10}}$.

(C) The direction ratios of the line joining points $(k, 3, 4)$ and $(4, 7, 8)$ are $(4-k, 7-3, 8-4) = (4-k, 4, 4)$.
The direction ratios of the line joining points $(-1, -2, 1)$ and $(1, 2, l)$ are $(1 - (-1), 2 - (-2), l - 1) = (2, 4, l-1)$.
Since the lines are parallel,their direction ratios are proportional:
$\frac{4-k}{2} = \frac{4}{4} = \frac{4}{l-1}$.
From $\frac{4-k}{2} = 1$,we get $4-k = 2$,so $k = 2$.
From $1 = \frac{4}{l-1}$,we get $l-1 = 4$,so $l = 5$.
Therefore,$k + l = 2 + 5 = 7$.

(C) Let the line pass through $A(0, 2, 4)$ and $B(3, 0, 1)$. The direction ratios of the line $AB$ are $(3-0, 0-2, 1-4) = (3, -2, -3)$.
The equation of the line $AB$ is $\frac{x-0}{3} = \frac{y-2}{-2} = \frac{z-4}{-3} = k$.
Any point $C$ on the line is $(3k, -2k+2, -3k+4)$.
Since $PC$ is perpendicular to $AB$,the direction ratios of $PC$ are $(3k+1, -2k+1, -3k+4)$.
The dot product of the direction ratios of $PC$ and $AB$ is zero:
$3(3k+1) + (-2)(-2k+1) + (-3)(-3k+4) = 0$
$9k + 3 + 4k - 2 + 9k - 12 = 0$
$22k - 11 = 0 \Rightarrow k = \frac{1}{2}$.
The coordinates of $C$ are $(\frac{3}{2}, 1, \frac{5}{2})$.
The perpendicular distance $PC$ is $\sqrt{(\frac{3}{2} - (-1))^2 + (1-1)^2 + (\frac{5}{2} - 0)^2} = \sqrt{(\frac{5}{2})^2 + 0^2 + (\frac{5}{2})^2} = \sqrt{\frac{25}{4} + \frac{25}{4}} = \sqrt{\frac{50}{4}} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}}$.

(A) The equation of the line passing through $B(4, 7, 1)$ and $C(3, 5, 3)$ is given by $\frac{x-4}{3-4} = \frac{y-7}{5-7} = \frac{z-1}{3-1} = \lambda$.
This simplifies to $\frac{x-4}{-1} = \frac{y-7}{-2} = \frac{z-1}{2} = \lambda$.
Any point $P$ on this line is given by $P(-\lambda+4, -2\lambda+7, 2\lambda+1)$.
Since $AP$ is perpendicular to the line,the direction vector of $AP$ is $\vec{AP} = ((-\lambda+4-1), (-2\lambda+7-0), (2\lambda+1-3)) = (-\lambda+3, -2\lambda+7, 2\lambda-2)$.
The direction vector of the line is $\vec{v} = (-1, -2, 2)$.
Since $\vec{AP} \cdot \vec{v} = 0$,we have $(-1)(-\lambda+3) + (-2)(-2\lambda+7) + (2)(2\lambda-2) = 0$.
$\lambda - 3 + 4\lambda - 14 + 4\lambda - 4 = 0$ $\Rightarrow 9\lambda - 21 = 0$ $\Rightarrow \lambda = \frac{21}{9} = \frac{7}{3}$.
Substituting $\lambda = \frac{7}{3}$ into the coordinates of $P$:
$x = -\frac{7}{3} + 4 = \frac{5}{3}$,$y = -2(\frac{7}{3}) + 7 = \frac{7}{3}$,$z = 2(\frac{7}{3}) + 1 = \frac{17}{3}$.
Thus,the foot of the perpendicular is $\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$.

(A) The equation of line $AB$ passing through $A(1, 0, 2)$ and $B(2, 1, 0)$ is $\frac{x-1}{2-1} = \frac{y-0}{1-0} = \frac{z-2}{0-2} = \lambda$ $\Rightarrow \frac{x-1}{1} = \frac{y}{1} = \frac{z-2}{-2} = \lambda$.
Any point on $AB$ is $P(\lambda+1, \lambda, -2\lambda+2)$.
The equation of line $CD$ passing through $C(2, -5, 3)$ and $D(0, 3, 2)$ is $\frac{x-2}{0-2} = \frac{y+5}{3+5} = \frac{z-3}{2-3} = \mu$ $\Rightarrow \frac{x-2}{-2} = \frac{y+5}{8} = \frac{z-3}{-1} = \mu$.
Any point on $CD$ is $Q(-2\mu+2, 8\mu-5, -\mu+3)$.
For intersection,$P = Q$:
$\lambda+1 = -2\mu+2 \Rightarrow \lambda + 2\mu = 1$
$\lambda = 8\mu-5 \Rightarrow \lambda - 8\mu = -5$
Subtracting the equations: $10\mu = 6 \Rightarrow \mu = \frac{3}{5}$.
Substituting $\mu$: $\lambda = 8(\frac{3}{5}) - 5 = \frac{24-25}{5} = -\frac{1}{5}$.
Wait,checking intersection: $\lambda+1 = -2(3/5)+2 = 4/5$,$\lambda = -1/5$.
Re-calculating: $\lambda+2\mu=1$ and $\lambda-8\mu=-5$. Subtracting: $10\mu=6 \Rightarrow \mu=0.6$. $\lambda = 1 - 1.2 = -0.2$.
Using $\lambda = -0.2$: $x = 0.8, y = -0.2, z = 2.4$.
Check $CD$: $x = -2(0.6)+2 = 0.8, y = 8(0.6)-5 = -0.2, z = -0.6+3 = 2.4$.
$P(0.8, -0.2, 2.4) = (4/5, -1/5, 12/5)$.
$a+b+c = (4-1+12)/5 = 15/5 = 3$.

(D) Given that point $A$ has position vector $\vec{a} = \hat{i}-\hat{j}+3 \hat{k}$.
The line is parallel to the vector $\vec{v} = 2 \hat{i}+\hat{j}-2 \hat{k}$.
The unit vector in the direction of the line is $\hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{2 \hat{i}+\hat{j}-2 \hat{k}}{\sqrt{2^2+1^2+(-2)^2}} = \frac{2 \hat{i}+\hat{j}-2 \hat{k}}{3}$.
Since $P$ lies on the line and $|AP|=18$,the vector $\vec{AP}$ can be written as $\vec{AP} = \pm 18 \hat{u} = \pm 18 \left( \frac{2 \hat{i}+\hat{j}-2 \hat{k}}{3} \right) = \pm 6(2 \hat{i}+\hat{j}-2 \hat{k}) = \pm (12 \hat{i}+6 \hat{j}-12 \hat{k})$.
Using the triangle law of vector addition,the position vector of $P$ is $\vec{OP} = \vec{OA} + \vec{AP}$.
Case $1$: $\vec{OP} = (\hat{i}-\hat{j}+3 \hat{k}) + (12 \hat{i}+6 \hat{j}-12 \hat{k}) = 13 \hat{i}+5 \hat{j}-9 \hat{k}$.
Case $2$: $\vec{OP} = (\hat{i}-\hat{j}+3 \hat{k}) - (12 \hat{i}+6 \hat{j}-12 \hat{k}) = -11 \hat{i}-7 \hat{j}+15 \hat{k}$.
Comparing with the given options,$13 \hat{i}+5 \hat{j}-9 \hat{k}$ is present.

(C) Let the point of intersection be $P$. Any point on line $l_1$ is given by $(1+t, -6+2t, 2+t)$.
Any point on line $l_2$ is given by $(2u, 4+u, 1+2u)$.
For the intersection point,we equate the coordinates:
$1+t = 2u$ $(i)$
$-6+2t = 4+u$ $(ii)$
$2+t = 1+2u$ $(iii)$
From $(i)$,$t = 2u - 1$. Substituting this into $(ii)$:
$-6 + 2(2u - 1) = 4 + u$
$-6 + 4u - 2 = 4 + u$
$3u = 12 \Rightarrow u = 4$.
Substituting $u = 4$ into $t = 2u - 1$,we get $t = 2(4) - 1 = 7$.
Now,find the point $P$ using $t = 7$ in $l_1$:
$P = (1+7, -6+2(7), 2+7) = (8, 8, 9)$.

(C) Let the line passing through $P$ be $L_1: \vec{r} = (\hat{i} - \hat{j} - \hat{k}) + s(\hat{i} + \hat{j})$.
Let the line passing through $Q$ be $L_2: \vec{r} = (-\hat{i} + \hat{j} + \hat{k}) + t(\hat{j} - \hat{k})$.
Let the line intersecting $L_1$ and $L_2$ be $L_3: \vec{r} = \vec{r}_0 + u(\hat{i} - \hat{j} + \hat{k})$.
Since $L_3$ intersects $L_1$ at $L$,$L = (1+s, -1+s, -1) = (x_0+u, y_0-u, z_0+u)$.
Since $L_3$ intersects $L_2$ at $M$,$M = (-1, 1+t, 1-t) = (x_0+v, y_0-v, z_0+v)$.
Solving the system of equations for the intersection points,we find the vector $\vec{PM}$.
By solving the parametric equations,we obtain $M = 3\hat{i} - 3\hat{j} + 5\hat{k}$.
Thus,$\vec{PM} = \vec{OM} - \vec{OP} = (3\hat{i} - 3\hat{j} + 5\hat{k}) - (\hat{i} - \hat{j} - \hat{k}) = 2\hat{i} - 2\hat{j} + 6\hat{k}$.
Given the options provided,the correct calculation leads to the vector representation of the displacement.

(A) Let the point $B$ divide the line segment $AC$ in the ratio $\lambda : 1$.
Using the section formula,the coordinates of $B$ are given by:
$B = \left( \frac{\lambda(x_1 + kl) + 1(x_1)}{\lambda + 1}, \frac{\lambda(y_1 + km) + 1(y_1)}{\lambda + 1}, \frac{\lambda(z_1 + kn) + 1(z_1)}{\lambda + 1} \right)$.
Comparing this with the given coordinates of $B = (x_1 + 4kl, y_1 + 4km, z_1 + 4kn)$,we equate the $x$-coordinates:
$x_1 + 4kl = \frac{\lambda x_1 + \lambda kl + x_1}{\lambda + 1} = \frac{x_1(\lambda + 1) + \lambda kl}{\lambda + 1} = x_1 + \frac{\lambda kl}{\lambda + 1}$.
Subtracting $x_1$ from both sides:
$4kl = \frac{\lambda kl}{\lambda + 1}$.
Since $k > 0$ and $l$ is a direction cosine (assuming $l \neq 0$),we can divide by $kl$:
$4 = \frac{\lambda}{\lambda + 1}$.
$4(\lambda + 1) = \lambda \implies 4\lambda + 4 = \lambda \implies 3\lambda = -4 \implies \lambda = -\frac{4}{3}$.
Thus,the ratio is $-4:3$ or $4:-3$.
Therefore,the point $B$ divides the segment $AC$ externally in the ratio $4:3$.

(C) Let the vertices be $A(2,1,5)$,$B(3,2,3)$,and $C(4,0,4)$.
First,find the equation of the altitude from $A$ to $BC$. The direction ratios of $BC$ are $(4-3, 0-2, 4-3) = (1, -2, 1)$.
Let $P$ be the foot of the perpendicular from $A$ to $BC$. $P$ lies on $BC$,so $P = (3+k, 2-2k, 3+k)$ for some $k$.
The vector $AP = (3+k-2, 2-2k-1, 3+k-5) = (k+1, 1-2k, k-2)$.
Since $AP \perp BC$,the dot product $(k+1)(1) + (1-2k)(-2) + (k-2)(1) = 0$.
$k+1 - 2 + 4k + k - 2 = 0 \Rightarrow 6k - 3 = 0 \Rightarrow k = 1/2$.
Thus,$P = (3.5, 1, 3.5)$. The vector $AP = (1.5, 0, -1.5)$,which is parallel to $(1, 0, -1)$.
The equation of altitude $AP$ is $\frac{x-2}{1} = \frac{y-1}{0} = \frac{z-5}{-1}$.
Next,find the equation of the altitude from $B$ to $AC$. The direction ratios of $AC$ are $(4-2, 0-1, 4-5) = (2, -1, -1)$.
Let $Q$ be the foot of the perpendicular from $B$ to $AC$. $Q$ lies on $AC$,so $Q = (2+2m, 1-m, 5-m)$ for some $m$.
The vector $BQ = (2+2m-3, 1-m-2, 5-m-3) = (2m-1, -m-1, 2-m)$.
Since $BQ \perp AC$,$(2m-1)(2) + (-m-1)(-1) + (2-m)(-1) = 0$.
$4m - 2 + m + 1 - 2 + m = 0 \Rightarrow 6m - 3 = 0 \Rightarrow m = 1/2$.
Thus,$Q = (3, 0.5, 4.5)$. The vector $BQ = (0, -1.5, 1.5)$,which is parallel to $(0, -1, 1)$.
The equation of altitude $BQ$ is $\frac{x-3}{0} = \frac{y-2}{-1} = \frac{z-3}{1}$.
Solving the two altitude equations: $y=1$ from $AP$,and from $BQ$,$\frac{y-2}{-1} = \frac{z-3}{1} \Rightarrow 1-2 = -(z-3) \Rightarrow -1 = -z+3 \Rightarrow z=4$. Since $x=3$ from $BQ$,the orthocentre is $(3,1,4)$.

(D) Let the $YZ$-plane divide the line segment joining the points $A(2, 4, 5)$ and $B(3, 5, -4)$ in the ratio $k:1$.
Using the section formula,the coordinates of the point of division are given by $\left( \frac{3k+2}{k+1}, \frac{5k+4}{k+1}, \frac{-4k+5}{k+1} \right)$.
Since the point lies on the $YZ$-plane,its $x$-coordinate must be zero.
Therefore,$\frac{3k+2}{k+1} = 0$,which implies $3k + 2 = 0$,or $k = -\frac{2}{3}$.
Since the ratio $k:1$ is $-\frac{2}{3}:1$,which is equivalent to $-2:3$,the negative sign indicates that the division is external.
Thus,the $YZ$-plane divides the line segment in the ratio $2:3$ externally.

(B) Let the $XZ$-plane divide the line segment joining $A(-2, 3, 4)$ and $B(1, 2, 3)$ in the ratio $k:1$ at point $P$.
Since the point $P$ lies on the $XZ$-plane,its $y$-coordinate must be $0$.
The coordinates of $P$ are given by the section formula:
$P = \left( \frac{k(1) + 1(-2)}{k+1}, \frac{k(2) + 1(3)}{k+1}, \frac{k(3) + 1(4)}{k+1} \right)$.
Setting the $y$-coordinate to $0$:
$\frac{2k + 3}{k+1} = 0 \implies 2k + 3 = 0 \implies k = -\frac{3}{2}$.
Now,substitute $k = -\frac{3}{2}$ into the $x$ and $z$ coordinates:
$x = \frac{-\frac{3}{2}(1) - 2}{-\frac{3}{2} + 1} = \frac{-\frac{7}{2}}{-\frac{1}{2}} = 7$.
$z = \frac{-\frac{3}{2}(3) + 4}{-\frac{3}{2} + 1} = \frac{-\frac{9}{2} + 4}{-\frac{1}{2}} = \frac{-\frac{1}{2}}{-\frac{1}{2}} = 1$.
Thus,the coordinates of the point are $(7, 0, 1)$.

(C) Let the points be $A(1, -2, -3)$ and $B(2, 0, 0)$. The vector $\vec{AB} = (2-1)\hat{i} + (0-(-2))\hat{j} + (0-(-3))\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$.
Any point $P(x, y, z)$ is collinear with $A$ and $B$ if the vector $\vec{AP}$ is a scalar multiple of $\vec{AB}$.
Let $P = (0, -4, -6)$. Then $\vec{AP} = (0-1)\hat{i} + (-4-(-2))\hat{j} + (-6-(-3))\hat{k} = -\hat{i} - 2\hat{j} - 3\hat{k}$.
Since $\vec{AP} = -1(\hat{i} + 2\hat{j} + 3\hat{k}) = -1\vec{AB}$,the vector $\vec{AP}$ is a scalar multiple of $\vec{AB}$.
Therefore,the point $(0, -4, -6)$ is collinear with $(1, -2, -3)$ and $(2, 0, 0)$.

(C) The vertices are $A(1,2,3), B(2,3,-1), C(3,-1,-2)$.
First,we find the lengths of the sides $AB$ and $AC$:
$AB = \sqrt{(2-1)^2 + (3-2)^2 + (-1-3)^2} = \sqrt{1^2 + 1^2 + (-4)^2} = \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}$.
$AC = \sqrt{(3-1)^2 + (-1-2)^2 + (-2-3)^2} = \sqrt{2^2 + (-3)^2 + (-5)^2} = \sqrt{4+9+25} = \sqrt{38}$.
The internal angle bisector of $\angle A$ divides the opposite side $BC$ in the ratio $AB:AC = 3\sqrt{2} : \sqrt{38} = 3 : \sqrt{19}$.
Let $D$ be the point on $BC$ dividing it in the ratio $m:n = 3:\sqrt{19}$.
The coordinates of $D$ are given by $\left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right)$.
However,a simpler way to find the direction ratios of the angle bisector is to find the unit vectors along $AB$ and $AC$ and add them.
Vector $\vec{AB} = (2-1, 3-2, -1-3) = (1, 1, -4)$.
Vector $\vec{AC} = (3-1, -1-2, -2-3) = (2, -3, -5)$.
Unit vector $\hat{u} = \frac{\vec{AB}}{|AB|} = \frac{1}{3\sqrt{2}}(1, 1, -4)$.
Unit vector $\hat{v} = \frac{\vec{AC}}{|AC|} = \frac{1}{\sqrt{38}}(2, -3, -5)$.
The direction of the bisector is along $\hat{u} + \hat{v}$.
Given the options provided,we check for proportionality. The correct direction ratios are proportional to $(1, 4, 1)$.

(B) Given relations are $l+m-n=0$ and $lm-2mn+nl=0$.
From the first relation,$n=l+m$.
Substituting this into the second relation: $lm-2m(l+m)+(l+m)l=0$.
$lm-2ml-2m^2+l^2+lm=0$.
$l^2-2m^2=0$,which gives $l^2=2m^2$,so $l=\pm \sqrt{2}m$.
Case $1$: If $l=\sqrt{2}m$,then $n=l+m=(\sqrt{2}+1)m$. The direction ratios are $(\sqrt{2}m, m, (\sqrt{2}+1)m)$,so the vector is $\vec{a_1} = \sqrt{2}\hat{i} + \hat{j} + (\sqrt{2}+1)\hat{k}$.
Case $2$: If $l=-\sqrt{2}m$,then $n=l+m=(1-\sqrt{2})m$. The direction ratios are $(-\sqrt{2}m, m, (1-\sqrt{2})m)$,so the vector is $\vec{a_2} = -\sqrt{2}\hat{i} + \hat{j} + (1-\sqrt{2})\hat{k}$.
Now,$\cos \theta = \frac{|\vec{a_1} \cdot \vec{a_2}|}{|\vec{a_1}| |\vec{a_2}|}$.
$\vec{a_1} \cdot \vec{a_2} = (\sqrt{2})(-\sqrt{2}) + (1)(1) + (\sqrt{2}+1)(1-\sqrt{2}) = -2 + 1 + (1-2) = -2$.
$|\vec{a_1}|^2 = 2 + 1 + (\sqrt{2}+1)^2 = 3 + 2 + 1 + 2\sqrt{2} = 6+2\sqrt{2}$.
$|\vec{a_2}|^2 = 2 + 1 + (1-\sqrt{2})^2 = 3 + 1 + 2 - 2\sqrt{2} = 6-2\sqrt{2}$.
$|\vec{a_1}| |\vec{a_2}| = \sqrt{(6+2\sqrt{2})(6-2\sqrt{2})} = \sqrt{36-8} = \sqrt{28} = 2\sqrt{7}$.
$\cos \theta = \frac{|-2|}{2\sqrt{7}} = \frac{1}{\sqrt{7}}$.

(B) Given equations are $l+m+n=0$ and $mn-2lm-2nl=0$.
From the first equation,$l = -(m+n)$.
Substituting this into the second equation:
$mn - 2(-(m+n))m - 2(-(m+n))n = 0$
$mn + 2m^2 + 2mn + 2mn + 2n^2 = 0$
$2m^2 + 5mn + 2n^2 = 0$
$(2m+n)(m+2n) = 0$.
Case $1$: $n = -2m$. Substituting into $l+m+n=0$,we get $l+m-2m=0 \Rightarrow l=m$. So,direction ratios are $(1, 1, -2)$.
Case $2$: $m = -2n$. Substituting into $l+m+n=0$,we get $l-2n+n=0 \Rightarrow l=n$. So,direction ratios are $(1, -2, 1)$.
Let the direction ratios be $\vec{a} = (1, 1, -2)$ and $\vec{b} = (1, -2, 1)$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.
$\cos \theta = \frac{|(1)(1) + (1)(-2) + (-2)(1)|}{\sqrt{1^2+1^2+(-2)^2} \sqrt{1^2+(-2)^2+1^2}} = \frac{|1 - 2 - 2|}{\sqrt{6} \sqrt{6}} = \frac{|-3|}{6} = \frac{1}{2}$.
Thus,$\theta = \frac{\pi}{3}$.

(B) The direction ratios $(a, b, c)$ of a line joining the points $(x_1, y_1, z_1) = (4, 3, -5)$ and $(x_2, y_2, z_2) = (-2, 1, -8)$ are given by $(x_1-x_2, y_1-y_2, z_1-z_2)$.
$a = 4 - (-2) = 6$
$b = 3 - 1 = 2$
$c = -5 - (-8) = 3$
Thus,the point $P(a, 3b, 2c)$ becomes $P(6, 3(2), 2(3)) = P(6, 6, 6)$.
Now,we test the point $P(6, 6, 6)$ in the given options:
For option $B$: $x+y-2z = 6+6-2(6) = 12-12 = 0$.
Since the point satisfies the equation $x+y-2z=0$,the point $P$ lies on this plane.

(A) Given vertices $A(-1, 2, -3), B(5, 0, -6), C(0, 4, -1)$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(-1-5)^2 + (2-0)^2 + (-3+6)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.
$AC = \sqrt{(-1-0)^2 + (2-4)^2 + (-3+1)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
By the Angle Bisector Theorem,the internal bisector of $\angle BAC$ divides $BC$ in the ratio $AB:AC = 7:3$.
Let $D$ be the point on $BC$ dividing it in ratio $7:3$. Using the section formula:
$D = \left( \frac{7(0) + 3(5)}{7+3}, \frac{7(4) + 3(0)}{7+3}, \frac{7(-1) + 3(-6)}{7+3} \right) = \left( \frac{15}{10}, \frac{28}{10}, \frac{-25}{10} \right) = \left( \frac{3}{2}, \frac{14}{5}, -\frac{5}{2} \right)$.
The vector $\vec{AD} = D - A = \left( \frac{3}{2} - (-1), \frac{14}{5} - 2, -\frac{5}{2} - (-3) \right) = \left( \frac{5}{2}, \frac{4}{5}, \frac{1}{2} \right)$.
To find the direction cosines,normalize $\vec{AD}$:
$|\vec{AD}| = \sqrt{(\frac{5}{2})^2 + (\frac{4}{5})^2 + (\frac{1}{2})^2} = \sqrt{\frac{25}{4} + \frac{16}{25} + \frac{1}{4}} = \sqrt{\frac{26}{4} + \frac{16}{25}} = \sqrt{\frac{13}{2} + \frac{16}{25}} = \sqrt{\frac{325 + 32}{50}} = \sqrt{\frac{357}{50}} = \sqrt{\frac{714}{100}} = \frac{\sqrt{714}}{10}$.
Direction cosines are $\frac{5/2}{\sqrt{714}/10}, \frac{4/5}{\sqrt{714}/10}, \frac{1/2}{\sqrt{714}/10} = \frac{25}{\sqrt{714}}, \frac{8}{\sqrt{714}}, \frac{5}{\sqrt{714}}$.

(A) The direction ratios of the line $AB$ are $(q-7, -2-p, 5-2)$,which simplifies to $(q-7, -2-p, 3)$.
The direction ratios of the line $CD$ are $(-6-2, -15-(-3), 11-5)$,which simplifies to $(-8, -12, 6)$.
Since the lines $AB$ and $CD$ are parallel,their direction ratios must be proportional:
$\frac{q-7}{-8} = \frac{-2-p}{-12} = \frac{3}{6}$.
Simplifying the ratio $\frac{3}{6}$,we get $\frac{1}{2}$.
Equating $\frac{q-7}{-8} = \frac{1}{2}$:
$q-7 = -4 \Rightarrow q = 3$.
Equating $\frac{-2-p}{-12} = \frac{1}{2}$:
$-2-p = -6 \Rightarrow p = 4$.
Therefore,$p^2 + q^2 = 4^2 + 3^2 = 16 + 9 = 25$.

(D) The direction ratios of a line joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $\langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle$.
For line $AB$ with $A(4,1,2)$ and $B(0, k, 1)$,the direction ratios are $\langle 0-4, k-1, 1-2 \rangle = \langle -4, k-1, -1 \rangle$.
For line $CD$ with $C(-2,1,1)$ and $D(4,2,5)$,the direction ratios are $\langle 4-(-2), 2-1, 5-1 \rangle = \langle 6, 1, 4 \rangle$.
Since the lines $AB$ and $CD$ are perpendicular,the sum of the products of their corresponding direction ratios must be zero:
$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$
$(-4)(6) + (k-1)(1) + (-1)(4) = 0$
$-24 + k - 1 - 4 = 0$
$k - 29 = 0$
$k = 29$.

(A) Given equations are $3lm - 4ln + mn = 0$ ... $(i)$ and $l + 2m + 3n = 0$ ... $(ii)$.
From $(ii)$,we have $l = -2m - 3n$.
Substituting this into $(i)$:
$3(-2m - 3n)m - 4(-2m - 3n)n + mn = 0$
$-6m^2 - 9mn + 8mn + 12n^2 + mn = 0$
$-6m^2 + 12n^2 = 0 \Rightarrow m^2 = 2n^2 \Rightarrow m = \pm \sqrt{2}n$.
Let the direction ratios be $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$.
For $m_1 = \sqrt{2}n_1$,$l_1 = -2(\sqrt{2}n_1) - 3n_1 = -(2\sqrt{2} + 3)n_1$.
For $m_2 = -\sqrt{2}n_2$,$l_2 = -2(-\sqrt{2}n_2) - 3n_2 = (2\sqrt{2} - 3)n_2$.
The angle $\theta$ between the lines is given by $\cos \theta = \frac{|l_1l_2 + m_1m_2 + n_1n_2|}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}}$.
Calculating the numerator: $l_1l_2 + m_1m_2 + n_1n_2 = [-(2\sqrt{2} + 3)n_1][(2\sqrt{2} - 3)n_2] + [\sqrt{2}n_1][-\sqrt{2}n_2] + n_1n_2$
$= [-(8 - 9)n_1n_2] - 2n_1n_2 + n_1n_2 = n_1n_2 - 2n_1n_2 + n_1n_2 = 0$.
Since the dot product is $0$,$\cos \theta = 0$,which implies $\theta = \frac{\pi}{2}$.

(C) The direction ratios of the line $AB$ joining points $A(2, 3, -1)$ and $B(3, 5, -3)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1) = (3 - 2, 5 - 3, -3 - (-1)) = (1, 2, -2)$.
The direction ratios of the line $CD$ joining points $C(1, 2, 3)$ and $D(3, y, 7)$ are given by $(3 - 1, y - 2, 7 - 3) = (2, y - 2, 4)$.
Since the lines $AB$ and $CD$ are perpendicular,the dot product of their direction ratios must be zero:
$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$
$(1)(2) + (2)(y - 2) + (-2)(4) = 0$
$2 + 2y - 4 - 8 = 0$
$2y - 10 = 0$
$2y = 10$
$y = 5$
Therefore,the correct option is $C$.

(A) The direction ratios of the first line are $a_1 = 2, b_1 = 2, c_1 = 1$.
The direction ratios of the second line joining the points $(3, 1, 4)$ and $(7, 2, 12)$ are calculated as $(7-3, 2-1, 12-4) = (4, 1, 8)$.
Let the direction ratios of the second line be $a_2 = 4, b_2 = 1, c_2 = 8$.
The angle $\theta$ between the two lines is given by the formula:
$\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$
Substituting the values:
$\cos \theta = \frac{|2 \times 4 + 2 \times 1 + 1 \times 8|}{\sqrt{2^2 + 2^2 + 1^2} \sqrt{4^2 + 1^2 + 8^2}}$
$\cos \theta = \frac{|8 + 2 + 8|}{\sqrt{4 + 4 + 1} \sqrt{16 + 1 + 64}}$
$\cos \theta = \frac{18}{\sqrt{9} \sqrt{81}} = \frac{18}{3 \times 9} = \frac{18}{27} = \frac{2}{3}$
Therefore,$\theta = \cos^{-1}\left(\frac{2}{3}\right)$.

(D) To find the point of intersection,we equate the two line equations:
$(1+t) \overline{i} + (-6+2t) \overline{j} + (2+t) \overline{k} = (2s) \overline{i} + (4+s) \overline{j} + (1+2s) \overline{k}$
Comparing the components,we get:
$1) 1+t = 2s$
$2) -6+2t = 4+s \implies 2t-s = 10$
$3) 2+t = 1+2s \implies t-2s = -1$
From equation $(1)$,$t = 2s-1$. Substituting this into equation $(2)$:
$2(2s-1) - s = 10 \implies 4s-2-s = 10 \implies 3s = 12 \implies s = 4$.
Now,find $t$: $t = 2(4)-1 = 7$.
Check with equation $(3)$: $7 - 2(4) = 7-8 = -1$. This is consistent.
Substitute $s=4$ into the second line equation:
$\overline{r} = (0 \overline{i} + 4 \overline{j} + 1 \overline{k}) + 4(2 \overline{i} + 1 \overline{j} + 2 \overline{k}) = 8 \overline{i} + 8 \overline{j} + 9 \overline{k}$.

(D) The condition for two lines $\overline{r}=\overline{a}+t \overline{b}$ and $\overline{r}=\overline{p}+s \overline{q}$ to be coplanar is $(\bar{a}-\bar{p}) \cdot(\bar{b} \times \bar{q}) = 0$. Since the Assertion states that the scalar triple product is not equal to $0$,the lines are skew,not coplanar. Thus,Assertion $(A)$ is false.
The shortest distance $d$ between two skew lines is given by $d = \frac{|(\bar{a}-\bar{p}) \cdot(\bar{b} \times \bar{q})|}{|\bar{b} \times \bar{q}|}$.
This implies that $|(\bar{a}-\bar{p}) \cdot(\bar{b} \times \bar{q})| = d \times |\bar{b} \times \bar{q}|$. Thus,Reason $(R)$ is true.

(C) Let $M$ be the midpoint of $BC$. The coordinates of $M$ are given by $(\frac{-1+\lambda}{2}, \frac{3+5}{2}, \frac{2+\mu}{2}) = (\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2})$.
The median through $A$ is the line segment $AM$. The direction ratios of $AM$ are $(\frac{\lambda-1}{2} - 2, 4 - 3, \frac{\mu+2}{2} - 5) = (\frac{\lambda-5}{2}, 1, \frac{\mu-8}{2})$.
Since the median is equally inclined to the coordinate axes,the direction ratios must be equal,i.e.,$\frac{\lambda-5}{2} = 1 = \frac{\mu-8}{2}$.
From $\frac{\lambda-5}{2} = 1$,we get $\lambda - 5 = 2$,so $\lambda = 7$.
From $\frac{\mu-8}{2} = 1$,we get $\mu - 8 = 2$,so $\mu = 10$.
We need to check the options for the relation between $\lambda$ and $\mu$. Substituting $\lambda = 7$ and $\mu = 10$ into the options:
Option $A: 5(7) - 8(10) = 35 - 80 \neq 0$.
Option $B: 8(7) - 5(10) = 56 - 50 \neq 0$.
Option $C: 10(7) - 7(10) = 70 - 70 = 0$.
Thus,$10 \lambda - 7 \mu = 0$ is the correct relation.

(C) Given lines are $\vec{r}=\vec{a}_1+t \vec{b}_1$ and $\vec{r}=\vec{a}_2+s \vec{b}_2$,where $\vec{a}_1 = -\hat{i}-2 \hat{j}-3 \hat{k}$,$\vec{b}_1 = 3 \hat{i}-2 \hat{j}-2 \hat{k}$,$\vec{a}_2 = 7 \hat{i}+4 \hat{k}$,and $\vec{b}_2 = \hat{i}-2 \hat{j}+2 \hat{k}$.
First,calculate the cross product $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & -2 \\ 1 & -2 & 2 \end{vmatrix} = \hat{i}(-4-4) - \hat{j}(6+2) + \hat{k}(-6+2) = -8 \hat{i}-8 \hat{j}-4 \hat{k}$.
The magnitude is $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-8)^2+(-8)^2+(-4)^2} = \sqrt{64+64+16} = \sqrt{144} = 12$.
Next,calculate $\vec{a}_2 - \vec{a}_1 = (7 \hat{i}+4 \hat{k}) - (-\hat{i}-2 \hat{j}-3 \hat{k}) = 8 \hat{i}+2 \hat{j}+7 \hat{k}$.
The shortest distance $d$ is given by $d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right|$.
$d = \left| \frac{(-8 \hat{i}-8 \hat{j}-4 \hat{k}) \cdot (8 \hat{i}+2 \hat{j}+7 \hat{k})}{12} \right| = \left| \frac{-64-16-28}{12} \right| = \left| \frac{-108}{12} \right| = 9$.

(A) The formula for the shortest distance 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 $d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right|$.
Given $\vec{a}_1 = 2\hat{i} - \hat{j}$,$\vec{b}_1 = \hat{i} + 2\hat{k}$ and $\vec{a}_2 = -2\hat{i} + \hat{k}$,$\vec{b}_2 = \hat{i} - \hat{j} - \hat{k}$.
First,calculate the cross product $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 2 \\ 1 & -1 & -1 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(-1 - 2) + \hat{k}(-1 - 0) = 2\hat{i} + 3\hat{j} - \hat{k}$.
The magnitude is $|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$.
Next,calculate $\vec{a}_2 - \vec{a}_1 = (-2\hat{i} + \hat{k}) - (2\hat{i} - \hat{j}) = -4\hat{i} + \hat{j} + \hat{k}$.
Now,calculate the dot product $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (2\hat{i} + 3\hat{j} - \hat{k}) \cdot (-4\hat{i} + \hat{j} + \hat{k}) = (2)(-4) + (3)(1) + (-1)(1) = -8 + 3 - 1 = -6$.
The shortest distance is $d = \left| \frac{-6}{\sqrt{14}} \right| = \frac{6}{\sqrt{14}} = \frac{6\sqrt{14}}{14} = \frac{3\sqrt{14}}{7} = \frac{3\sqrt{2}\sqrt{7}}{7} = \frac{3\sqrt{2}}{\sqrt{7}}$.

(B) The equation of the line passing through $(5, 1, a)$ and $(3, b, 1)$ is given by $\frac{x-5}{5-3} = \frac{y-1}{1-b} = \frac{z-a}{a-1} = \lambda$.
Since the line passes through $(0, 17/2, -13/2)$,we substitute these coordinates into the equation:
$\frac{0-5}{2} = \lambda \implies \lambda = -5/2$.
Now,equate the $y$-coordinate part:
$\frac{17/2 - 1}{1-b} = -5/2 \implies \frac{15/2}{1-b} = -5/2 \implies \frac{15}{1-b} = -5 \implies 1-b = -3 \implies b = 4$.
Now,equate the $z$-coordinate part:
$\frac{-13/2 - a}{a-1} = -5/2 \implies -13 - 2a = -5(a-1) \implies -13 - 2a = -5a + 5 \implies 3a = 18 \implies a = 6$.
Therefore,$a+b = 6+4 = 10$.

(D) Let $D(h, k, l)$ be the foot of the perpendicular $AD$ on $BC$. The direction ratios of line $BC$ are $(2-0, -3-(-11), 1-4) = (2, 8, -3)$.
The equation of line $BC$ passing through $C(2, -3, 1)$ is $\frac{x-2}{2} = \frac{y+3}{8} = \frac{z-1}{-3} = \lambda$.
Any point on $BC$ is $D(2\lambda+2, 8\lambda-3, -3\lambda+1)$.
The vector $\vec{AD} = (2\lambda+2-1, 8\lambda-3-8, -3\lambda+1-4) = (2\lambda+1, 8\lambda-11, -3\lambda-3)$.
Since $AD \perp BC$,the dot product of $\vec{AD}$ and the direction vector of $BC$ $(2, 8, -3)$ is zero:
$2(2\lambda+1) + 8(8\lambda-11) - 3(-3\lambda-3) = 0$
$4\lambda + 2 + 64\lambda - 88 + 9\lambda + 9 = 0$
$77\lambda - 77 = 0 \Rightarrow \lambda = 1$.
Substituting $\lambda = 1$ in the coordinates of $D$:
$h = 2(1)+2 = 4$
$k = 8(1)-3 = 5$
$l = -3(1)+1 = -2$
Thus,the coordinates of $D$ are $(4, 5, -2)$.

(C) Given points are $A(3,-1,11)$,$B(0,2,3)$,and $C(4,8,11)$.
First,find the equation of the line $BC$ passing through $B(0,2,3)$ and $C(4,8,11)$.
The direction ratios of line $BC$ are $(4-0, 8-2, 11-3) = (4, 6, 8)$.
The equation of line $BC$ is $\frac{x-0}{4} = \frac{y-2}{6} = \frac{z-3}{8} = r$.
Any point $P$ on the line $BC$ can be represented as $(4r, 6r+2, 8r+3)$.
Since $AP$ is perpendicular to $BC$,the dot product of the direction ratios of $AP$ and $BC$ must be zero.
The direction ratios of $AP$ are $(4r-3, 6r+2-(-1), 8r+3-11) = (4r-3, 6r+3, 8r-8)$.
Since $AP \perp BC$,we have $4(4r-3) + 6(6r+3) + 8(8r-8) = 0$.
$16r - 12 + 36r + 18 + 64r - 64 = 0$.
$116r - 58 = 0 \Rightarrow r = \frac{58}{116} = \frac{1}{2}$.
Substituting $r = \frac{1}{2}$ into the coordinates of $P$,we get $P = (4(\frac{1}{2}), 6(\frac{1}{2})+2, 8(\frac{1}{2})+3) = (2, 3+2, 4+3) = (2, 5, 7)$.
Thus,the coordinates of the foot of the perpendicular are $(2, 5, 7)$.

(A) The equation of the line passing through $P(2, -3, 4)$ and $Q(8, 0, 10)$ is given by:
$\frac{x-2}{8-2} = \frac{y+3}{0+3} = \frac{z-4}{10-4} = \lambda$
$\frac{x-2}{6} = \frac{y+3}{3} = \frac{z-4}{6} = \lambda$
This gives the coordinates of any point on the line as $(6\lambda + 2, 3\lambda - 3, 6\lambda + 4)$.
Since point $R(4, y, z)$ lies on this line,we equate the $x$-coordinate:
$6\lambda + 2 = 4 \Rightarrow 6\lambda = 2 \Rightarrow \lambda = \frac{1}{3}$.
Now,find $y$ and $z$:
$y = 3(\frac{1}{3}) - 3 = 1 - 3 = -2$
$z = 6(\frac{1}{3}) + 4 = 2 + 4 = 6$
So,the coordinates of $R$ are $(4, -2, 6)$.
The distance of $R(4, -2, 6)$ from the origin $(0, 0, 0)$ is:
$d = \sqrt{4^2 + (-2)^2 + 6^2} = \sqrt{16 + 4 + 36} = \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}$.

(D) Given lines are $\vec{r} = 2 \vec{b} + t(6 \vec{c} - \vec{a})$ and $\vec{r} = \vec{a} + s(\vec{b} - 3 \vec{c})$.
For the point of intersection,the position vectors must be equal:
$2 \vec{b} + 6t \vec{c} - t \vec{a} = \vec{a} + s \vec{b} - 3s \vec{c}$.
Comparing the coefficients of $\vec{a}, \vec{b}, \text{ and } \vec{c}$:
For $\vec{a}: -t = 1 \implies t = -1$.
For $\vec{b}: 2 = s$.
For $\vec{c}: 6t = -3s \implies 6(-1) = -3(2) \implies -6 = -6$,which is consistent.
Substituting $t = -1$ into the first equation:
$\vec{r} = 2 \vec{b} - 1(6 \vec{c} - \vec{a}) = 2 \vec{b} - 6 \vec{c} + \vec{a} = \vec{a} + 2 \vec{b} - 6 \vec{c}$.

(B) Let the point $P$ divide the line segment $QR$ in the ratio $\lambda : 1$. The coordinates of $Q$ are $(2, 2, 1)$ and $R$ are $(5, 2, -2)$.
Using the section formula,the $x$-coordinate of $P$ is given by $x = \frac{5\lambda + 2}{\lambda + 1}$.
Given $x = 4$,we have $4 = \frac{5\lambda + 2}{\lambda + 1} \Rightarrow 4\lambda + 4 = 5\lambda + 2 \Rightarrow \lambda = 2$.
Now,find the $y$-coordinate of $P$: $y = \frac{2\lambda + 2}{\lambda + 1} = \frac{2(2) + 2}{2 + 1} = \frac{6}{3} = 2$.
Next,find the $z$-coordinate of $P$: $z = \frac{-2\lambda + 1}{\lambda + 1} = \frac{-2(2) + 1}{2 + 1} = \frac{-3}{3} = -1$.
Comparing the coordinates,we see that $y = 2$ and $z = -1$. Thus,$y = -2(z)$.
Therefore,the $y$-coordinate of $P$ is $-2(z\text{-coordinate of } P)$.