The direction cosines of the line of intersection of the planes $x+2y+z-4=0$ and $2x-y+z-3=0$ are

The distance of the point $(2, 3, 4)$ from the plane $3x - 6y + 2z + 11 = 0$ is

If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$,then $(\alpha, \beta)=$

The point $\bar{i}-2 \bar{j}$ lies on a line parallel to the vector $2 \bar{i}+\bar{k}$. The point $\bar{i}+2 \bar{j}$ lies on a plane parallel to the vectors $2 \bar{j}-\bar{k}$ and $\bar{i}+2 \bar{k}$. Find the point of intersection of the line and the plane.

The coordinates of the foot of the perpendicular from the point $(1, -2, 1)$ on the plane containing the lines $\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}$ and $\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}$ 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
$(A) \frac{2}{\sqrt{75}}$ $(B) \frac{7}{\sqrt{75}}$ $(C) \frac{13}{\sqrt{75}}$ $(D) \frac{23}{\sqrt{75}}$