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 $c = ....$

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 $\vec{r}=(s+1) \hat{i}+(2 s-1) \hat{j}-(2 s+1) \hat{k}$.

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 . . . . . . .

The shortest distance between the lines $\frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5}$ and $\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3}$ is

The shortest distance between the lines ${r_1} = 4i - 3j - k + \lambda (i - 4j + 7k)$ and ${r_2} = i - j - 10k + \mu (2i - 3j + 8k)$ is

Let $P$ be the foot of the perpendicular from the point $Q(10,-3,-1)$ on the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}$. Then the area of the right-angled triangle $PQR$,where $R$ is the point $(3,-2,1)$,is