Given that \( AD \) is the bisector of \( \angle BAC \), the angle bisector theorem states:
\[
\frac{BD}{DC} = \frac{AB}{AC} \quad {...(i)}
\]
Next, we compute the lengths of \( AB \) and \( AC \).
Calculation of \( AB \)
\[
AB = \sqrt{(5 - 3)^2 + (3 - 2)^2 + (2 - 0)^2}
\]
\[
= \sqrt{4 + 1 + 4} = \sqrt{9} = 3
\]
Calculation of \( AC \)
\[
AC = \sqrt{(-9 - 3)^2 + (6 - 2)^2 + (-3 - 0)^2}
\]
\[
= \sqrt{144 + 16 + 9}
\]
\[
= \sqrt{169} = 13
\]
Determining the Ratio \( BD : DC \)
Using equation (i):
\[
\frac{BD}{DC} = \frac{3}{13}
\]
Therefore, \( D \) divides \( BC \) in the ratio \( 3:13 \).
Determining the Coordinates of \( D \)
Employing the section formula:
\[
D \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right)
\]
With \( B(-9,6,-3) \) and \( C(5,3,2) \), and the ratio \( m:n = 3:13 \):
\[
x = \frac{3(-9) + 13(5)}{3+13}, \quad y = \frac{3(6) + 13(3)}{3+13}, \quad z = \frac{3(-3) + 13(2)}{3+13}
\]
\[
= \left( \frac{-27 + 65}{16}, \frac{18 + 39}{16}, \frac{-9 + 26}{16} \right)
\]
\[
= \left( \frac{38}{16}, \frac{57}{16}, \frac{17}{16} \right)
\]
\[
= \left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right)
\]
The coordinates of \( D \) are:
\[
D \left( \frac{19}{8}, \frac{57}{16}, \frac{17}{16} \right)
\]