The equation of the plane passing through the point $(1,1,1)$ and through the line of intersection of $x+2y-z+1=0$ and $3x-y-4z+3=0$ is

The distance of the point having position vector $\hat{i}-2 \hat{j}-6 \hat{k}$ from the straight line passing through the point $(2, -3, -4)$ and parallel to the vector $6 \hat{i}+3 \hat{j}-4 \hat{k}$ is units.

If $4x + 4y - kz = 0$ is the equation of the plane through the origin that contains the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4},$ then $k =$

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 equation of the plane passing through the origin and perpendicular to the line $x = 2y = 3z$ is

The Cartesian equation of a line passing through $(1, 2, 3)$ and parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$ is: