3. As shown in Figure 2, in $\triangle A B C$, $\angle B$ is a right angle, the angle bisector of $\angle A$ is $A D$, and the median on side $B C$ is $A E$. Points $D$ and $E$ divide $B C$ into three segments in the ratio $1: 2: 3$. Then $\sin \angle B A C=(\quad)$.
(A) $\frac{12}{13}$
(B) $\frac{4 \sqrt{3}}{9}$
(C) $\frac{2 \sqrt{6}}{5}$
(D) $\frac{2+\sqrt{3}}{4}$
3.C.

Let $B D=x, A B=y$, then $D E=2 x, E C=3 x$.
By $\frac{B D}{D C}=\frac{A B}{A C}$, we get $A C=5 y$.
Also, $A B^{2}+B C^{2}=A C^{2}$, which means $y^{2}+(6 x)^{2}=(5 y)^{2}$.
Therefore, $x^{2}=\frac{2 y^{2}}{3}, \sin \angle B A C=\frac{6 x}{5 y}=\frac{2 \sqrt{6}}{5}$.