A ( - 1 , 2 , - 3 ) ,   B ( 5 , 0 , - 6 ) ,   C ( 0 , 4 , - 1 ) are the vertices of a triangle A B…

Given points$A(-1,2,-3), B(5,0,-6), C(0,4,-1)$

Let, the vector $AD$is the internal bisector of $\angle BAC$, point $D$ lie on the side $BC$.

Now, $AB=\sqrt{{\left(-1-5\right)}^2+2^2+{\left(-3+6\right)}^2}=\sqrt{36+4+9}=7$

and, $AC=\sqrt{{\left(-1-0\right)}^2+{\left(2-4\right)}^2+{\left(-3+1\right)}^2}=\sqrt{1+4+4}=3$

We know that, the bisector of angle $\angle BAC$, divides the side $BC$ in ratio $AB:AC$, i.e. $7 : 3$

So, the coordinate of $D$ is $\left(\frac{3\times 5+7\times 0}{3+7}, \frac{3\times 0+7\times 4}{3+7}, \frac{3\times \left(-6\right)+7\times \left(-1\right)}{3+7}\right)=\left(\frac{3}{2}, \frac{14}{5}, -\frac{5}{2}\right)$

Hence, the vector $\overrightarrow{AD}=\frac{5}{2}\hat{i}+\frac{4}{5}\hat{j}+\frac{1}{2}\hat{k}$

So, the direction cosines are $\frac{{\displaystyle \frac{5}{2}}}{\sqrt{{\left({\displaystyle \frac{5}{2}}\right)}^2+{\left({\displaystyle \frac{4}{5}}\right)}^2+{\left({\displaystyle \frac{1}{2}}\right)}^2}}, \frac{{\displaystyle \frac{4}{5}}}{\sqrt{{\left({\displaystyle \frac{5}{2}}\right)}^2+{\left({\displaystyle \frac{4}{5}}\right)}^2+{\left({\displaystyle \frac{1}{2}}\right)}^2}}\text{'}\frac{{\displaystyle \frac{1}{2}}}{\sqrt{{\left({\displaystyle \frac{5}{2}}\right)}^2+{\left({\displaystyle \frac{4}{5}}\right)}^2+{\left({\displaystyle \frac{1}{2}}\right)}^2}}$

$\mathrm{i}.\mathrm{e}. \frac{{\displaystyle 25}}{\sqrt{714}}, \frac{{\displaystyle 8}}{\sqrt{714}},\frac{{\displaystyle 5}}{\sqrt{714}}$