Example 5. Solve the system of equations in the real number range:
$$
\left(\begin{array}{l}
a \sqrt{x-y+z} \cdot \sqrt{x+y-z}=x \sqrt{y z}, \\
b \sqrt{x+y-z} \cdot \sqrt{-x+y+z}=y \sqrt{z x}, \\
c \sqrt{-x+y+z} \cdot \sqrt{x-y+z}=z \sqrt{x y},
\end{array}\right.
$$

where $a, b, c \neq 0$.
If $x y z=0$, then at least one of $x, y, z$ is 0. In this case, the solution to the system of equations is obvious, i.e., $x=0$,
$$
\begin{array}{l}
y=z, y=0, z=x, z=0, x=y . \\
\quad \text { If } x y z \neq 0, \quad \text { then } x \neq 0, y \neq 0, z \neq 0 . \\
\because x y>0, y z>0, z x>0, x+y>z, \\
z+x>y, y+z>x, \\
\therefore x>0, y>0, z>0 .
\end{array}
$$

At this point, it only holds when $a>0, b>0, c>0$.

From the above, $x, y, z$ can serve as the three sides of $\triangle X Y Z$ (sides $x, y, z$ are opposite to angles $X, Y, Z$ respectively). Let $x+y+z=2 p$, by the half-angle formula, we get
$$
\begin{aligned}
\sin \frac{X}{2} & =\sqrt{\frac{(p-y)(\not-z)}{y z}} \\
& =\frac{1}{2} \sqrt{\frac{(x-y+z)(x+\dot{y}-\bar{z}}{y z}}, \\
\sin \frac{Y}{2} & =\frac{1}{2} \sqrt{\frac{(x+y-z)(-x+y+z)}{z x}}, \\
\sin \frac{Z}{2} & =\frac{1}{2} \sqrt{\frac{(-x+y+z)(x-y+z)}{x y}} .
\end{aligned}
$$

The system of equations can be transformed into
$$
\begin{array}{l}
2 a \sin \frac{X}{2}=x, \quad 2 b \sin \frac{Y}{2}=y, \\
2 c \sin \frac{Z}{2}=z .
\end{array}
$$

For the circumradius of $\triangle X Y Z$, by the sine rule, we get
$$
\begin{array}{l}
2 a \sin \frac{X}{2}=2 R \sin X, \\
2 b \sin \frac{Y}{2}=2 R \sin Y, \\
2 c \sin \frac{Z}{2}=2 R \sin Z, \\
a=2 R \cos \frac{X}{2}=2 R \sin \left(\frac{\pi}{2}-\frac{X}{2}\right), \\
b=2 R \cos \frac{Y}{2}=2 R \sin \left(\frac{\pi}{2}-\frac{Y}{2}\right), \\
c=2 R \cos \frac{Z}{2}=2 R \sin \left(\frac{\pi}{2}-\frac{Z}{2}\right) .
\end{array}
$$

Let $A=\frac{\pi}{2}-\frac{X}{2}, B=\frac{\pi}{2}-\frac{Y}{2}, C=\frac{\dot{\pi}}{2}-\frac{Z}{2}$, then
$$
\begin{array}{l}
a=2 R \sin A, \quad b=2 R \sin B, \\
c=2 R \sin C .
\end{array}
$$

At this point, it only holds when $a, b, c$ can serve as the sides of some acute triangle.
Since $A, B, C$ are all positive angles, and
$$
\begin{array}{l}
A+B+C=\left(\frac{\pi}{2}-\frac{X}{2}\right)+\left(\frac{\pi}{2}-\frac{Y}{2}\right) \\
+\left(\frac{\pi}{2}-\frac{Z}{2}\right) \\
=\frac{3 \pi}{2}-\frac{X+Y+Z}{2}=\pi, \\
\end{array}
$$
$\therefore$ From (2), $a, b, c$ are the sides opposite to angles $A, B, C$ of $\triangle A B C$. Thus,
$$
\left\{\begin{array}{l}
\cos \left(\frac{\pi}{2}-\frac{X}{2}\right)=\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}, \\
\cos \left(\frac{\pi}{2}-\frac{Y}{2}\right)=\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}, \\
\cos \left(\frac{\pi}{2}-\frac{Z}{2}\right)=\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}, \\
\text { then } \\
\sin \frac{X}{2}=\frac{b^{2}+c^{2}-a^{2}}{2 b c}, \\
\sin \frac{Y}{2}=\frac{c^{2}+a^{2}-b^{2}}{2 c a}, \\
\sin \frac{Z}{2}=\frac{a^{2}+b^{2}-c^{2}}{2 a b} .
\end{array}\right.
$$

Thus, from (1), we get
$$
\begin{array}{l}
x=2 a \sin \frac{X}{2}=\frac{a\left(b^{2}+c^{2}-a^{2}\right)}{b c}, \\
y=2 b \sin \frac{Y}{2}=\frac{b\left(c^{2}+a^{2}-b^{2}\right)}{c a}, \\
z=2 c \sin \frac{Z}{2}=\frac{c\left(a^{2}+b^{2}-c^{2}\right)}{a b} .
\end{array}
$$