Consider a $\triangle ABC$ where $A(1,3,2)$, $B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$, then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is:\\
(1) $\frac{37}{2\sqrt{38}}$\\
(2) $\frac{\sqrt{38}}{2}$\\
(3) $\frac{39}{2\sqrt{38}}$\\
(4) $\sqrt{19}$