The shortest distance between the lines $r = (3t - 4)\hat{i} - 2\hat{j} - (1 + 2t)\hat{k}$ and $r = (6 + s)\hat{i} + (2 - 2s)\hat{j} + 2(1 + s)\hat{k}$ is

The distance of the point $P(-3, 2, 3)$ from the line passing through $A(4, 6, -2)$ with direction ratios $\langle -1, 2, 3 \rangle$ is . . . . . . units.

The vector equation of the line $2x+4=3y+1=6z-3$ is

The points $A(3, 2, 0)$,$B(5, 3, 2)$,and $C(-9, 6, -3)$ are the vertices of a triangle $ABC$. If $AD$ is the internal bisector of $\angle BAC$ which meets $BC$ at $D$,then the coordinates of $D$ are:

Let $P$ be the image of the point $Q(7,-2,5)$ in the line $L: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $R(5, p, q)$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is $\qquad$

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