Example 2 Try to find a solution to the system of equations
$$
\begin{array}{l}
\frac{x\left(y^{2}+z^{2}\right)}{a \sqrt{p}}=\frac{y\left(z^{2}+x^{2}\right)}{b \sqrt{q}}=\frac{z\left(x^{2}+y^{2}\right)}{c \sqrt{r}} \\
=\sqrt{p q r}
\end{array}
$$

where $a, b, c, p, q, r$ are positive numbers satisfying
$$
\frac{s-a}{p}=\frac{s-b}{q}=\frac{s-c}{r}, s=\frac{1}{2}(a+b+c) \text {. }
$$
Let $x=\sqrt{q r} u, y=\sqrt{p p} v, z=\sqrt{p q} w$, then the original system of equations transforms to
$$
\begin{array}{l}
\frac{u\left(r v^{2}+q w^{2}\right)}{a}=\frac{v\left(p w^{2}+r u^{2}\right)}{b} \\
=\frac{w\left(q u^{2}+p v^{2}\right)}{c}=1 .
\end{array}
$$

If $s-a \leqslant 0$, then $s-b \leqslant 0, s-c \leqslant 0$. Adding these three inequalities gives $s \leqslant 0$, leading to a contradiction. Therefore, $s-a>0$. Similarly, $s-b>0, s-c>0$. Thus, $a, b, c$ can serve as the three sides of a triangle $ABC$. Construct three circles with centers at $A, B, C$ and each pair of circles externally tangent to each other. It is evident that the radii are $s-a, s-b, s-c$. Since $(s-b)+(s-c)=a, (s-c)+(s-a)=b, (s-a)+(s-b)=c$, if the system of equations
$$
\left\{\begin{array}{l}
p v^{2} w=p v w^{2}=s-a, \\
q w^{2} u=q w u^{2}=s-b, \\
r u^{2} v=r w w^{2}=s-c
\end{array}\right.
$$

has a solution $(u, v, w)$, then this solution is also a solution to system (1). Let $\frac{s-a}{p}=\frac{s-b}{q}=\frac{s-c}{r}=k^{3}$, then the solution to (2) is $u=v=w=k$.
Therefore, the original system of equations has a solution
$$
(x, y, z)=(k \sqrt{q r}, k \sqrt{p}, k \sqrt{p q}) .
$$