If A(-1,2,-3), B(5,0,-6) and C(0,4,-1) are the vertices of $∆$ABC, then direction ratio’s of the external bisector of $\mathrm{\angle }\mathrm{BAC}$ are:

If the external bisector of $\mathrm{\angle }\mathrm{BAC}$ meets BC at E, then E divides BC externally in the ratio of AB:AC.
AB = 7, AC = 3
$\therefore$Coordinates of E are $\left(\frac{-15}{4},7,\frac{11}{4}\right)$
$\therefore$The direction ratio of the line AE are $(-\frac{15}{4}+1,7-2,\frac{11}{4}+3)=(\frac{-11}{4},5,\frac{23}{4})=(-11,20,23)$