If $A(0,3,4), B(1,5,6), C(-2,0,-2)$ are the vertices of a triangle $ABC$ and the bisector of angle $A$ meets the side $BC$ at $D$, then $AD=$

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$ is $\frac{5}{\sqrt{6}}$,then the sum of all possible values of $\alpha$ is

The square of the distance of the point $\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ from the line $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector $\hat{i}+4 \hat{j}+7 \hat{k}$ is :

The shortest distance between the skew-lines $\frac{x-3}{-1}=\frac{y-4}{2}=\frac{z+2}{1}$ and $\frac{x-1}{1}=\frac{y+7}{3}=\frac{z+2}{2}$ is

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

$A$ line passing through the points $P(1, -1, 2)$ and $Q(2, 0, 1)$ meets the $XY$-plane and the $YZ$-plane at points $A$ and $B$ respectively. The distance $AB$ is equal to...