If the foot of the perpendicular from point $(4,3,8)$ on the line $L_{1}: \frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4},$ $l \neq 0$ is $(3,5,7),$ then the shortest distance between the line $L_{1}$ and line $L_{2}: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is equal to:

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

The acute angle between the line joining the points $(2,1,-3)$ and $(-3,1,7)$ and a line parallel to $\frac{x-1}{3}=\frac{y}{4}=\frac{z+3}{5}$ is

The equation of a line passing through the point $(2, -1, 1)$ and parallel to the line joining the points $\hat{i} + 2\hat{j} + 2\hat{k}$ and $-\hat{i} + 4\hat{j} + \hat{k}$ 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 . . . . . . .

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