The line joining the points $(3, 5, -7)$ and $(-2, 1, 8)$ meets the $yz$-plane at point

If the distances of the point $P(1, 2, a)$ from the line $L: \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $L_{1}: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $L_{2}: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal,then $a+b+c$ is equal to

The angle between a line with direction ratios $2, 2, 1$ and the line joining the points $(3, 1, 4)$ and $(7, 2, 12)$ is

Find the perpendicular distance from the point $P(4, -5, 3)$ to the line $\vec{r} = (5, -2, 6) + k(3, -4, 5)$,where $k \in \mathbb{R}$.

The two lines $L_1: \vec{r}=(\hat{i}+5 \hat{j}+5 \hat{k})+t(4 \hat{i}-4 \hat{j}+5 \hat{k})$ and $L_2: \vec{r}=(2 \hat{i}+4 \hat{j}+5 \hat{k})+s(8 \hat{i}-3 \hat{j}+\hat{k})$ are such that

The shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$ is