The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are at right angles,then $k =$

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

Let $P$ be the foot of the perpendicular from the point $A(1, 2, 2)$ on the line $L: \frac{x-1}{1} = \frac{y+1}{-1} = \frac{z-2}{2}$. Let the line $\overrightarrow{r} = (-\hat{i} + \hat{j} - 2\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$,$\lambda \in R$,intersect the line $L$ at $Q$. Then $2(PQ)^2$ is equal to:

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other,then $k = \dots$

Find the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$.