10 In the tetrahedron $S-A B C$, $S A=4, S B \geqslant 7, S C \geqslant 9, A B=5, B C \leqslant$ $6, A C \leqslant 8$. Then the maximum volume of the tetrahedron $S-A B C$ is $\qquad$.
10 Let $\angle S A B=\alpha$, according to the cosine theorem we have
$$
\begin{aligned}
\cos \alpha & =\frac{S A^{2}+A B^{2}-S B^{2}}{2 \times S A \times A B} \\
& \leqslant \frac{4^{2}+5^{2}-7^{2}}{2 \times 4 \times 5} \\
& =-\frac{1}{5},
\end{aligned}
$$

Therefore,
$$
\begin{aligned}
\sin \alpha & =\sqrt{1-\cos ^{2} \alpha} \leqslant \frac{2 \sqrt{6}}{5}, \\
S_{\triangle S A B} & =\frac{1}{2} \times S A \times A B \times \sin \alpha \leqslant 4 \sqrt{6} .
\end{aligned}
$$

Since the height of the pyramid does not exceed its slant height, we have
$$
V_{C-S A B} \leqslant \frac{1}{3} S_{\triangle S A B} \times B C \leqslant 8 \sqrt{6} .
$$

In fact, when $S B=7, B C=6$ and $C B \perp$ plane $S A B$, it can be verified that the given conditions are satisfied, at this time $V_{S-A B C}=8 \sqrt{6}$, the volume of the pyramid can reach the maximum value of $8 \sqrt{6}$.