A(3,2,0),B(5,3,2),C(-9,6,-3)are three points forming a triangle then the coordinates of a point in which the bisector of $\angle BAC$ meets BC is

$$\begin{gathered} Given points A(3,2,0),B(5,3,2),C(-9,6,-3) \\ Now AB=\sqrt{{(5-3)}^2+{(3-2)}^2+{(2-0)}^2} \\ =\sqrt{9}=3 \\ and AC=\sqrt{{(-9-5)}^2+{(6-3)}^2+{(-3-2)}^2} \\ =\sqrt{169}=13 \\ D divides BC in the ratio 3:13 internally \\ Now D=(\frac{3(-9)+13\left(5\right)}{16},\frac{3\left(6\right)+13\left(3\right)}{16},\frac{3(-3)+13\left(2\right)}{16}) \\ =(\frac{38}{16},\frac{57}{16},\frac{17}{16}) \end{gathered}$$