7. A regular tetrahedron frame with edge length 3 contains a solid sphere with radius 1. Then the maximum volume of the common part of the tetrahedron and the sphere is $\qquad$
7. $\left(\frac{7 \sqrt{6}}{8}-\frac{4}{3}\right) \pi$.

From the problem, using the formula for the volume of a spherical cap, we know that the volume of a spherical cap of a sphere with radius $r$ cut by a plane at a distance $d (0<d<r)$ from the center of the sphere is
$$
\left(\frac{1}{3} d^{3}-r^{2} d+\frac{2}{3} r^{3}\right) \pi
$$

Let the distances from the center $O$ of the sphere to the four faces of the regular tetrahedron $ABCD$ be $d_{1}, d_{2}, d_{3}, d_{4}$.

Since the sphere intersects all four faces of the tetrahedron and the center of the sphere is inside the tetrahedron, we have
$$
\begin{array}{l}
V_{\text {tetrahedron } O-BCD}+V_{\text {tetrahedron } O-CDA}+ \\
V_{\text {tetrahedron } O-DAB}+V_{\text {tetrahedron } O-ABC} \\
=V_{\text {tetrahedron } A-BCD} \\
\Rightarrow d_{1}+d_{2}+d_{3}+d_{4} \\
=\text { the distance from point } A \text { to plane } BCD \\
=\sqrt{6} \text {. } \\
\end{array}
$$

Then the volume of the intersecting part is
$$
\begin{array}{l}
V_{0}=V_{\text {sphere }}-\sum_{i=1}^{4}\left(\frac{1}{3} d_{i}^{3}-r^{2} d_{i}+\frac{2}{3} r^{3}\right) \pi \\
=\frac{4 \pi}{3}-\frac{\pi}{3} \sum_{i=1}^{4} d_{i}^{3}+\sqrt{6} \pi-\frac{8 \pi}{3} \text { (since } r=1 \text { ) } \\
\leqslant\left(\sqrt{6}-\frac{4}{3}\right) \pi \times \frac{\pi}{3} \cdot \frac{\left(\sum^{4}+\frac{\pi}{d_{i}}\right)^{23}:}{16} \\
\end{array}
$$
(Arithmetic Mean Inequality)
$$
=\left(\frac{7 \sqrt{6}}{8}-\frac{4}{3}\right) \pi,
$$

with equality holding if and only if the center of the sphere is at the center of the regular tetrahedron.