The square of the distance of the point $(-2, -8, 6)$ from the line $\frac{x-1}{1} = \frac{y-1}{2} = \frac{z}{-1}$ along the line $\frac{x+5}{1} = \frac{y+5}{-1} = \frac{z}{2}$ is equal to:

Let the line $L$ pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$. Then,which of the following points lies on the line $L$?

If the lines $\frac{x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$ and $\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ are perpendicular,find the value of $k$.

Let $Q$ and $R$ be two points on the line $\frac{x+1}{2} = \frac{y+2}{3} = \frac{z-1}{2}$ at a distance of $\sqrt{26}$ from the point $P(4, 2, 7)$. Then the square of the area of the triangle $PQR$ is $....$

Let $l_{1}$ be the line in $xy$-plane with $x$ and $y$ intercepts $\frac{1}{8}$ and $\frac{1}{4 \sqrt{2}}$ respectively,and $l_{2}$ be the line in $zx$-plane with $x$ and $z$ intercepts $-\frac{1}{8}$ and $-\frac{1}{6 \sqrt{3}}$ respectively. If $d$ is the shortest distance between the line $l_{1}$ and $l_{2}$,then $d^{-2}$ is equal to

The length of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$ is: