The shortest distance between the skew lines $\vec{r}=(2 \hat{i}-\hat{j})+t(\hat{i}+2 \hat{k})$ and $\vec{r}=(-2 \hat{i}+\hat{k})+s(\hat{i}-\hat{j}-\hat{k})$ is

Find the vector and the Cartesian equations of the line passing through the point $(5, 2, -4)$ and parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$.

Let a line passing through the point $(-1, 2, 3)$ intersect the lines $L_1: \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2}$ at $M(\alpha, \beta, \gamma)$ and $L_2: \frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4}$ at $N(a, b, c)$. Then the value of $\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$ equals

If the lines $\frac{x - 1}{c} = \frac{y + 2}{-2} = \frac{z - 3}{4}$ and $\frac{x - 5}{1} = \frac{y - 3}{1} = \frac{z + 1}{c}$ are parallel,then $c = ....$

The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6}$ and $\frac{x-3}{2} = \frac{y-3}{3} = \frac{z+5}{6}$ is . . . . . . .

Find the shortest distance between the lines whose vector equations are $\vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k}$ and $\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$.