The equation of the plane containing the lines $\frac{x - 1}{2} = \frac{y + 1}{\lambda} = \frac{z}{2}$ and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{\lambda}$ is

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ is

Let $L_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$ and $L_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$ be two lines. Let $L_3$ be a line passing through the point $(\alpha, \beta, \gamma)$ and be perpendicular to both $L_1$ and $L_2$. If $L_3$ intersects $L_1$,then $|5\alpha-11\beta-8\gamma|$ equals :

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
$(A) \frac{2}{\sqrt{75}}$ $(B) \frac{7}{\sqrt{75}}$ $(C) \frac{13}{\sqrt{75}}$ $(D) \frac{23}{\sqrt{75}}$

The foot of the perpendicular drawn from a point $A(1,1,1)$ onto a plane $\pi$ is $P(-3,3,5)$. If the equation of the plane parallel to the plane $\pi$ and passing through the midpoint of $AP$ is $ax-y+cz+d=0$,then $a+c-d=$

The distance of the point $P(3,4,4)$ from the point of intersection of the line joining the points $Q(3,-4,-5)$ and $R(2,-3,1)$ with the plane $2x+y+z=7$ is: (in $units$)