7. As shown in the figure, in rectangle $A B C D$, $A B=3, A D=4, E$ is a point on $A B$, and $A E=1$. Now, $\triangle B C E$ is folded along $C E$ so that the projection of point $B$ on plane $A E C D$ falls on $A D$. Then, the volume of the pyramid $B-A E C D$ is $\qquad$ .
Draw $B F \perp E C$, intersecting $A D, E C$ at $F, G$ respectively.
It is easy to find that $B F=\frac{3 \sqrt{5}}{2}, B G=\frac{4 \sqrt{5}}{5}$.
By the projection theorem, we have
$$
\begin{array}{l}
B^{\prime} F^{2}=\frac{3 \sqrt{5}}{2} \cdot\left(\frac{8 \sqrt{5}}{5}-\frac{3 \sqrt{5}}{2}\right)=\frac{3}{4} \\
\Rightarrow B^{\prime} F=\frac{\sqrt{3}}{2} .
\end{array}
$$

Therefore, $V_{B-A E C D}=\frac{1}{3} \cdot \frac{1}{2} \cdot 4 \cdot 4 \cdot \frac{\sqrt{3}}{2}=\frac{4 \sqrt{3}}{3}$.