If the lines $x = ay - 1 = z - 2$ and $x = 3y - 2 = bz - 2$ $(ab \neq 0)$ are coplanar,then

$A$ perpendicular is drawn from a point $P$ on the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z}{1}$ to the plane $x + y + z = 3$ such that the foot of the perpendicular $Q$ also lies on the plane $x - y + z = 3$. Then the coordinates of $Q$ are

$A$ line with positive direction cosines passes through the point $P(2,1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ equals $\qquad$ units.

$X$ intercept of the plane containing the line of intersection of the planes $x-2y+z+2=0$ and $3x-y-z+1=0$ and also passing through $(1,1,1)$ is

Consider the lines $L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}$.
$1.$ The unit vector perpendicular to both $L_1$ and $L_2$ is
$(A) \frac{-\hat{i}+7 \hat{j}+7 \hat{k}}{\sqrt{99}}$ $(B) \frac{-\hat{i}-7 \hat{j}+5 \hat{k}}{5 \sqrt{3}}$ $(C) \frac{-\hat{i}+7 \hat{j}+5 \hat{k}}{5 \sqrt{3}}$ $(D) \frac{7 \hat{i}-7 \hat{j}-\hat{k}}{\sqrt{99}}$
$2.$ The shortest distance between $L_1$ and $L_2$ is
$(A) 0$ $(B) \frac{17}{\sqrt{3}}$ $(C) \frac{41}{5 \sqrt{3}}$ $(D) \frac{17}{5 \sqrt{3}}$
$3.$ The distance of the point $(1,1,1)$ from the plane passing through the point $(-1,-2,-1)$ and whose normal is perpendicular to both the lines $L_1$ and $L_2$ is
$(A) \frac{2}{\sqrt{75}}$ $(B) \frac{7}{\sqrt{75}}$ $(C) \frac{13}{\sqrt{75}}$ $(D) \frac{23}{\sqrt{75}}$

If $L$ is a line common to the planes $3x + 4y + 7z = 1$ and $x - y + z = 5$, then the direction ratios of the line $L$ are: