For a positive real number $p$,if the perpendicular distance from a point $-\hat{i} + p\hat{j} - 3\hat{k}$ to the plane $\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 7$ is $6$ units,then $p=$

The direction cosines of the line formed by the intersection of the planes $x - y + 2z = 5$ and $3x + y + z = 6$ are:

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

Let the mirror image of the point $(a, b, c)$ with respect to the plane $3x - 4y + 12z + 19 = 0$ be $(a - 6, \beta, \gamma)$. If $a + b + c = 5$,then $7\beta - 9\gamma$ is equal to

Find the vector equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane $\vec{r} \cdot (\hat{i} + 2\hat{j} - 5\hat{k}) + 9 = 0$.

The equation of the plane passing through the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4}$ and perpendicular to the plane $x+2y+z=12$ is given by $ax+by+cz+4=0$. Then: