If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect,then $k = . . . . .$.

Find the foot of the perpendicular from the point $(2, 3, -8)$ to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$. Also,find the perpendicular distance from the given point to the line.

If the distance of the point $(a, 2, 5)$ from the image of the point $(1, 2, 7)$ in the line $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-2}{2}$ is $4$, then the sum of all possible values of $a$ is equal to :

Lines $\frac{2x-5}{k} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other. Then,the value of $k$ is . . . . . . .

If the direction cosines of two lines are given by $l+m+n=0$ and $mn-2lm-2nl=0$,then the acute angle between those lines is

$A$ 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: