If $\mathrm{A}\left(1, -1, -3\right), \mathrm{B}\left(2, 1, -2\right), \mathrm{C}\left(-5, 2, -6\right)$ are the vertices of a $∆\mathrm{ABC}$, then the length of internal  bisector of angle A is

$$\begin{gathered} \mathrm{AB}=\sqrt{1+4+1}=\sqrt{6} \\ \mathrm{AC}=\sqrt{36+9+9}=3\sqrt{6} \\ \mathrm{AB}:\mathrm{AC}=1:3 \\ \overline{\mathrm{OD}}=\frac{-5\mathrm{j}+2\mathrm{j}-6\mathrm{k}+6\mathrm{i}+3\mathrm{j}-6\mathrm{k}}{4}=\frac{\mathrm{i}+5\mathrm{j}-12\mathrm{k}}{4} \\ \overline{\mathrm{Ad}}=\frac{\mathrm{i}+5\mathrm{j}-12\mathrm{k}}{4}-(\mathrm{i}-\mathrm{j}-3\mathrm{k}) \\ =\frac{-3\mathrm{i}+9\mathrm{j}}{4} \\ \left|\overline{\mathrm{AD}}\right|=\frac{\sqrt{9+81}}{4}=\frac{3\sqrt{10}}{4} \end{gathered}$$