3. Given in $\triangle A B C$, $A B=2, A C=3, \angle B A C=60^{\circ}, \overrightarrow{C D}=2 \overrightarrow{B E}, \overrightarrow{A E}=x \overrightarrow{A D}+$ $(1-x) \overrightarrow{A B}, x \in(0,1)$. Then the range of the projection of $A E$ on $A C$ is $\qquad$
3. $(1,7)$.

Taking $A$ as the origin, the direction of $\overrightarrow{A C}$ as the positive direction of the $x$-axis to establish a Cartesian coordinate system, then $C(3,0), B(1, \sqrt{3})$, $D(7,-2 \sqrt{3})$, then $\overrightarrow{A E}=(6 x+1,-3 \sqrt{3} x+\sqrt{3}),|\overrightarrow{A E}| \cos \angle E A C=\frac{\overrightarrow{A E} \cdot \overrightarrow{A C}}{|\overrightarrow{A C}|}=\frac{3(6 x+1)}{3}=$ $6 x+1$. Given $x \in(0,1)$, we know $6 x+1 \in(1,7)$.