$\text{ Let }A(3,2,0),B(5,3,2),C(-9,6,-3)\text{ are three points forming a triangle. If }AD\text{ , the }$$\text{ bisector of }\mathrm{\angle }BAC\text{ meets }BC\text{ in }D,\text{ then coordinates of }D\text{ are }$

$A = \left(3,2,0 \right), B=( 5,3,2) , C \left(-9,6,-3\right)$

$AB = \sqrt{4+1+4} =3 , AC =\sqrt{144+16+9 }=13$

AD = angular bisector  of $\angle BAC$

 since  BD : DC = 3 :13   $\Rightarrow D =\left(\frac{-27+65}{16},\frac{18+39}{16}\text{'}\frac{-9+26}{16}\right),$    by section formula

                                          $= \left(\frac{38}{16},\frac{57}{16},\frac{17}{16}\right)$    =$\left(\frac{19}{8},\frac{57}{16},\frac{17}{16}\right)$