The lines $\frac{x-3}{1}=\frac{y-2}{1}=\frac{z-5}{-k}$ and $\frac{x-4}{k}=\frac{y-3}{1}=\frac{z-3}{2}$ are coplanar,hence $k=$

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

If the foot of the perpendicular drawn from the point $A(1, 0, 3)$ on a line passing through $B(\alpha, 7, 1)$ is $P\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$,then $\alpha$ is equal to:

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

Find the shortest distance between the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2\hat{i} - \hat{j} - \hat{k})$ and $\vec{r} = (\hat{i} - \hat{j} - \hat{k}) + \mu(\hat{i})$.

Let $A(\alpha, 4, 7)$ and $B(3, \beta, 8)$ be two points in space. If the $YZ$ plane and $ZX$ plane respectively divide the line segment joining the points $A$ and $B$ in the ratio $2:3$ and $4:5$, then the point $C$ which divides $\overline{AB}$ in the ratio $\alpha: \beta$ externally is