(10 Given that the three lateral edges of a triangular pyramid are pairwise perpendicular, and the base area is 1, find the maximum volume of this triangular pyramid.



保留源文本的换行和格式，翻译结果如下：

(10 Given that the three lateral edges of a triangular pyramid are pairwise perpendicular, and the base area is 1, find the maximum volume of this triangular pyramid.
10 As shown in the figure, in the tetrahedron $P-ABC$, $PA$, $PB$, and $PC$ are mutually perpendicular. Let the dihedral angles between the faces $APB$, $BPC$, $APC$, and the base $ABC$ be $\alpha$, $\beta$, and $\gamma$ respectively. Given that $S_{\triangle ABC}=1$ and $PC \perp$ plane $APB$, the projection of $\triangle ABC$ onto plane $APB$ is $\triangle APB$. By the projection area formula, we have $\cos \alpha=\frac{S_{\triangle APB}}{S_{\triangle ABC}}$, which means $S_{\triangle APB}=S_{\triangle ABC} \cos \alpha=\cos \alpha$. Similarly, $S_{\triangle BPC}=\cos \beta$, $S_{\triangle APC}=\cos \gamma$.

Let the projection of point $P$ onto the base $ABC$ be $O$, and connect $AO$, $BO$, and $CO$. Since $PA$, $PB$, and $PC$ are mutually perpendicular, it is easy to prove that $\triangle ABC$ must be an acute triangle and $O$ is the orthocenter of $\triangle ABC$, so point $O$ must be inside $\triangle ABC$. By the projection area formula, we have $\cos \alpha=\frac{S_{\triangle AOB}}{S_{\triangle APB}}$, which means $S_{\triangle AOB}=S_{\triangle APB} \cos \alpha=\cos^2 \alpha$. Similarly, $S_{\triangle BOC}=\cos^2 \beta$, $S_{\triangle AOC}=\cos^2 \gamma$, and
$$
\begin{aligned}
1 & =S_{\triangle ABC} \\
& =S_{\triangle AOB}+S_{\triangle BOC}+S_{\triangle AOC} \\
& =\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma .
\end{aligned}
$$

We now find the volume.
$$
S_{\triangle APB}=\frac{1}{2} PA \cdot PB=\cos \alpha,
$$

so $PA \cdot PB=2 \cos \alpha$.
Similarly, $PB \cdot PC=2 \cos \beta$, $PA \cdot PC=2 \cos \gamma$.
Multiplying these three equations, we get
$$
PA \cdot PB \cdot PC=2 \sqrt{2 \cos \alpha \cos \beta \cos \gamma},
$$

thus
$$
V=\frac{1}{6} PA \cdot PB \cdot PC=\frac{1}{3} \sqrt{2 \cos \alpha \cos \beta \cos \gamma},
$$

By the AM-GM inequality $\frac{x+y+z}{3} \geqslant \sqrt[3]{xyz}$, we have
$$
\frac{\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma}{3} \geqslant \sqrt[3]{\cos^2 \alpha \cos^2 \beta \cos^2 \gamma},
$$

which means
$$
\cos^2 \alpha \cos^2 \beta \cos^2 \gamma \leqslant \left(\frac{1}{3}\right)^3=\frac{1}{27}.
$$

Thus, $\sqrt{\cos \alpha \cos \beta \cos \gamma} \leqslant \sqrt[4]{\frac{1}{27}}$, so $V \leqslant \frac{1}{3} \sqrt[4]{\frac{4}{27}}$, with equality when $\cos \alpha=\cos \beta=\cos \gamma=\frac{\sqrt{3}}{3}$, i.e., $\alpha=\beta=\gamma=\arccos \frac{\sqrt{3}}{3}$.