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

The Cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{6-z}{2}$. Find the vector equation of the line.

If the shortest distance between the lines $\frac{x-k}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$,then $k=$

The lines $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$

Let the line $L_{1}$ be parallel to the vector $-3\hat{i}+2\hat{j}+4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_{2}$ be parallel to the vector $2\hat{i}+\hat{j}+3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_{3}$ is parallel to the vector $-3\hat{i}+5\hat{j}+16\hat{k}$ and intersects the lines $L_{1}$ and $L_{2}$ at the points $C$ and $D$, respectively, then $|\overrightarrow{CD}|^2$ is equal to:

Find the angle between the lines whose direction cosines are given by the equations $3l+m+5n=0$ and $6mn-2nl+5lm=0$.