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

Let $A(2,3,-1), B(4,1,0), C(-1,-1,11)$ be the vertices of a triangle $ABC$. Let $D$ be the point where the bisector of $\angle BAC$ meets the side $BC$. Then the direction ratios of $AD$ are

If the foot of the perpendicular from point $(4,3,8)$ on the line $L_{1}: \frac{x-a}{l}=\frac{y-2}{3}=\frac{z-b}{4},$ $l \neq 0$ is $(3,5,7),$ then the shortest distance between the line $L_{1}$ and line $L_{2}: \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is equal to:

The coordinates of the point where the line joining the points $(3, 5, -7)$ and $(-2, 1, 8)$ is intersected by the $YOZ$ plane are

The length of the perpendicular from the point $(1, -2, 5)$ to the line passing through $(1, 2, 4)$ and parallel to the line $x + y - z = 0 = x - 2y + 3z - 5$ is:

If the line passing through the points $(-5, 1, 3)$ and $(1, 2, 0)$ is perpendicular to the line passing through the points $(x, 2, 1)$ and $(0, -4, 6)$,then $x = \dots$