(3) In the tetrahedron $P-ABC$, $\angle APB=\angle BPC=\angle CPA=90^{\circ}$, and the sum of the lengths of all edges is $S$. Then the maximum volume of this tetrahedron is $\qquad$.
(3) $\frac{S^{3}}{162(1+\sqrt{2})^{3}}$ Hint: When the edges $A P=B P=$
$C P$ of the tetrahedron $P-A B C$ are equal, the volume of this tetrahedron is maximized.
Let
$$
A P=B P=C P=x,
$$

then
$$
A B=B C=C A=\sqrt{2} x .
$$

From the given information,
$$
A P+B P+C P+A B+B C+C A=3(1+\sqrt{2}) x=S,
$$

thus $x=\frac{S}{3(1+\sqrt{2})}$. The maximum volume of the tetrahedron $P-A B C$ is $\frac{1}{6} x^{3}=$
$$
\frac{S^{3}}{162(1+\sqrt{2})^{3}} \text {. }
$$