The vertices of a triangle ABC

The given co-ordinates: A(3, 2, 0), B(5, 3, 2) and C(-9, 6, -3)

Now, $A B=\sqrt{(5-3)^{2}+(3-2)^{2}+(2-0)^{2}}=\sqrt{4+1+4}=3$

Also, $A C=\sqrt{(-9-3)^{2}+(6-2)^{2}+(-3-0)^{2}}=\sqrt{144+16+9}=13$

Now, we have, $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3}{13}$

By the property of internal angle bisector,

$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{CD}}$

Therefore, $\frac{\mathrm{BD}}{\mathrm{CD}}=\frac{3}{13}$

Applying the section formula, we get,

$D(x, y, z)=\left(\frac{3 \times 5-9 \times 13}{3+13}, \frac{3 \times 3+6 \times 13}{3+13}, \frac{3 \times 2-3 \times 13}{3+13}\right)$

$D(x, y, z)=\left(-\frac{102}{16}, \frac{87}{16}, \frac{33}{16}\right)$