7. In tetrahedron $A B C D$, $\angle A D B=\angle B D C$ $=\angle C D A=60^{\circ}, A D=B D=3, C D=2$. Then the volume of the circumscribed sphere of tetrahedron $A B C D$ is $\qquad$ .
7. $4 \sqrt{3} \pi$.

It is easy to know that $\triangle A D B$ is an equilateral triangle, and $C A = C B$. As shown in Figure 2, let $P$ and $M$ be the midpoints of $A B$ and $C D$ respectively, and connect $P D$ and $P C$. Then the plane $A B D \perp$ plane $P D C$.

Let $N$ be the circumcenter of $\triangle A B D$, and $O$ be the center of the circumscribed sphere of the tetrahedron $A B C D$. Then $O M \perp C D, O N \perp P D$.
It can be found that $\cos \angle C D P = \frac{\sqrt{3}}{3}, \sin \angle C D P = \frac{\sqrt{6}}{3}$.
By the problem, $D N = \sqrt{3}, D M = 1$.
In $\triangle D M N$, by the cosine rule we get
$$
\begin{array}{l}
M N^{2} = D M^{2} + D N^{2} - 2 D M \cdot D N \cos \angle C D P = 2 \\
\Rightarrow M N = \sqrt{2} .
\end{array}
$$

Since $D, M, O, N$ are four points on the circle with $D O$ as the diameter, we have
$$
D O = \frac{M N}{\sin \angle C D P} = \sqrt{3} .
$$

Therefore, the volume of the circumscribed sphere $V = \frac{4 \pi}{3} (\sqrt{3})^{3} = 4 \sqrt{3} \pi$.