Find the shortest distance between lines $\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and $\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})$. (in $units$)

Find the vector equation of the line $\frac{x - 2}{2} = \frac{2y - 5}{-3}, z = -1$.

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.)

Let a line $L$ be perpendicular to both the lines $L_1: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$ and $L_2: \frac{x-2}{1} = \frac{y-4}{4} = \frac{z-6}{7}$. If $\theta$ is the acute angle between the lines $L$ and $L_3: \frac{x-7}{2} = \frac{y-7}{1} = \frac{z}{2}$, then $\tan \theta$ is equal to:

The distance of the point $B(1, 2, 3)$ from the line passing through $A(4, 2, 2)$ and parallel to the vector $\vec{c} = 2i + 3j + 6k$ is

If $P$ is a point on the line parallel to the vector $2 \hat{i}-3 \hat{j}-6 \hat{k}$ and passing through the point $A$ whose position vector is $\hat{i}+2 \hat{j}-2 \hat{k}$ and $AP=21$, then the position vector of $P$ can be