$A$ line drawn from a point $A(-2,-2,3)$ and parallel to the line $\frac{x}{-2}=\frac{y}{2}=\frac{z}{-1}$ meets the $YOZ-$ plane in point $P$. Then,the coordinates of the point $P$ are:

The Cartesian equation of the line passing through the point $(5, -2, 4)$ and parallel to the vector $3\hat{i}-2\hat{j}+8\hat{k}$ is . . . . . . .

The vector equation of the line $2x+4=3y+1=6z-3$ is

The angle between the lines $2x = 3y = -z$ and $6x = -y = -4z$ is (in $^{\circ}$)

Let $O$ be the origin,and $M$ and $N$ be the points on the lines $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ and $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ respectively such that $MN$ is the shortest distance between the given lines. Then $\overrightarrow{OM} \cdot \overrightarrow{ON}$ is equal to:

The equation of the line,passing through $A(1, 2, 3)$ and perpendicular to the vectors $2 \hat{i} + \hat{j} - \hat{k}$ and $\hat{i} + 3 \hat{j} + 2 \hat{k}$,is