Example 5 The edge lengths of the tetrahedron $S-A B C$ are $S C=A B=$ $a, S A=B C=b, S B=C A=c$. Try to calculate the volume of this tetrahedron $S-A B C$.
Solution: As shown in Figure 15 of Example 4, we can deduce:
$$
\begin{array}{l}
x^{2}+z^{2}=4 c^{2}, \\
x^{2}+y^{2}=4 b^{2}, \\
y^{2}+z^{2}=4 a^{2} .
\end{array}
$$

Then, by adding and simplifying, we get
$$
x^{2}+y^{2}+z^{2}=2\left(a^{2}+b^{2}+c^{2}\right) \text {. }
$$
(4) - (3) yields
$$
x^{2}=2\left(b^{2}+c^{2}-a^{2}\right) \text {, }
$$
(4) - (1) yields
$$
y^{2}=2\left(a^{2}+b^{2}-c^{2}\right) \text {, }
$$
(4) - (2) yields
$$
\begin{array}{c}
z^{2}=2\left(c^{2}+a^{2}-b^{2}\right) . \\
\therefore x^{2} y^{2} z^{2}=8\left(a^{2}+b^{2}-c^{2}\right)\left(b^{2}+c^{2}\right. \\
\left.-a^{2}\right)\left(c^{2}+a^{2}-b^{2}\right), \\
x y z=2 \sqrt{2\left(a^{2}+b^{2}-c^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)\left(c^{2}+a^{2}-b^{2}\right)} . \\
\therefore V_{S-A B C}=\frac{1}{4} V_{S-s_{1} s_{2} s_{3}}=\frac{1}{4} \cdot \frac{1}{6} x y z \\
=\frac{1}{24} x y z \\
=\frac{1}{12} \sqrt{2\left(a^{2}+b^{2}-c^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)\left(c^{2}+a^{2}-b^{2}\right)} .
\end{array}
$$

This is the formula for calculating the volume of an isosceles tetrahedron based on the lengths of its sides.