$A=\left(-1,2,-3\right),B=\left(5,0,-6\right),C=\left(0,4,-1\right)$. If $l,m,n$ be the direction cosines of the internal angular bisector of$\angle BAC$, then $l-m-n=$

The vector along the internal angular bisector of $\overline{a},\overline{b}$ is$\left(\frac{\overline{a}}{\left|a\right|}\pm \frac{\overline{b}}{\left|b\right|}\right)$

Here$\overline{a}=\overline{AB}\text{\hspace{0.17em}and }\overline{b}=\overline{AC}$

It implies that $\overline{a}=6i-2j-3k\text{\hspace{0.17em}and }\overline{b}=i+2j+2k$

Hence, the vector along the internal angular bisector of  $\angle BAC$ is $\left(\frac{\overline{a}}{\left|a\right|}+\frac{\overline{b}}{\left|b\right|}\right)=\frac{25i+8j+5k}{21}$

Therefore, the direction cosines are$〈\frac{25}{\sqrt{714}},\frac{8}{\sqrt{714}},\frac{5}{\sqrt{714}}〉$

It implies that  $l-m-n=\boxed{\frac{12}{\sqrt{714}}}$