7. In the tetrahedron $S-ABC$,
$$
SA=SB=SC=\sqrt{21}, BC=6 \text{. }
$$

If the projection of point $A$ onto the plane of the side $SBC$ is exactly the orthocenter of $\triangle SBC$, then the volume of the inscribed sphere of the tetrahedron $S-ABC$ is . $\qquad$
7. $\frac{4 \pi}{3}$.

Since the projection of point $A$ on the plane containing side $SBC$ is exactly the orthocenter of $\triangle SBC$, we have
$$
S A \perp B C, S B \perp A C, S C \perp A B .
$$

Thus, the projection of point $S$ on the plane $ABC$ is the orthocenter of $\triangle ABC$.

Given $\sqrt{S A}=\sqrt{S B}=\sqrt{S C}=\sqrt{21}$, the projection of point $S$ on the plane $ABC$ is the circumcenter of $\triangle ABC$.
Therefore, $A B=B C=C A=6$.
Hence, $V_{\text {tetrahedron } S-A B C}=9 \sqrt{3}, S_{\text {tetrahedron } S-A B C}=27 \sqrt{3}$, and the inradius is $r=1$.
Thus, the volume of the inscribed sphere is $\frac{4 \pi}{3}$.