$A\left(-1,2,-3\right),B\left(5,0,-6\right)$ and $C\left(0,4,-1\right)$ are the vertices of a triangle. The d.r’s of the internal bisector of $\angle BAC$  are

$Given vertices are A=(-1,2,-3),B=(5,0,-6) and C=(0,4,-1)$

Now D.c’s BA are $\left(\frac{6}{7},\frac{-2}{7},\frac{-3}{7}\right)$

DC’s CA are $\left(\frac{1}{3},\frac{2}{3},\frac{2}{3}\right)$

$\therefore$ D.r ’s of internal bisector of $\lfloor BAC$ are 

$\left(\frac{6}{7}+\frac{1}{3},\frac{-2}{7}+\frac{2}{3},\frac{-3}{7}+\frac{2}{3}\right)=\left(\frac{25}{21},\frac{8}{21},\frac{5}{21}\right)=(25,8,5)$