Solve $a=BC=3\sqrt{6}, b=AC=3, c=AB=5$, thus $\cos A=\frac{9+25-54}{2 \cdot 3 \cdot 5}=-\frac{2}{3}$, so $\tan \frac{A}{2}=\frac{\sin A}{1+\cos A}=\sqrt{5}$, hence the answer is $A$.
3. In the spatial rectangular coordinate system, the three vertices of $\triangle A B C$ have coordinates $A(3,4,1), B(0,4,5), C(5,2,0)$, then the value of $\tan \frac{A}{2}$ is
A. $\sqrt{5}$
B. $\frac{\sqrt{5}}{5}$
C. $\sqrt{6}$
D. $\frac{\sqrt{6}}{6}$