Let $(\alpha, \beta, \gamma)$ be the mirror image of the point $(2, 3, 5)$ in the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$. Then $2\alpha + 3\beta + 4\gamma$ is equal to

The lines $\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 3}{0}$ and $\frac{x - 2}{0} = \frac{y - 3}{0} = \frac{z - 4}{1}$ are:

Find the coordinates of the point where the line passing through $(5, 1, 6)$ and $(3, 4, 1)$ crosses the $ZX$-plane.

Find the angle between the following pair of lines:
$\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}$ and $\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$

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

Let the line of the shortest distance between the lines $L_1: \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $L_2: \vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})$ intersect $L_1$ and $L_2$ at $P$ and $Q$ respectively. If $(\alpha, \beta, \gamma)$ is the midpoint of the line segment $PQ$,then $2(\alpha+\beta+\gamma)$ is equal to . . . . . . .