Let in a $\triangle ABC$,the length of the side $AC$ be $6$,the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ 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

Find the equation of the line which passes through the point $(1, 2, 3)$ and is parallel to the vector $3 \hat{i} + 2 \hat{j} - 2 \hat{k}$.

The distance between a point $P$ whose position vector is $5 \hat{i}+\hat{j}+3 \hat{k}$ and the line $r=(3 \hat{i}+7 \hat{j}+\hat{k})+t(\hat{j}+\hat{k})$ is

If $p$ is the shortest distance between the lines $\frac{x + 1}{7} = \frac{y + 1}{-6} = \frac{z + 1}{1}$ and $\vec{r} = (3\hat{i} + 5\hat{j} + 7\hat{k}) + \mu(\hat{i} - 2\hat{j} + \hat{k})$, then $[p]$ is... (where $[.]$ denotes the greatest integer function.)

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 . . . . . . .