$4 \cdot 130$ Solve the system of equations
$$
\left\{\begin{array}{l}
x^{2}=a+(y-z)^{2}, \\
y^{2}=b+(z-x)^{2}, \\
z^{2}=c+(x-y)^{2} .
\end{array}\right.
$$
[Solution] Move all variables in the system of equations to the left and factorize, we get
$$
\left\{\begin{array}{l}
(x-y+z)(x+y-z)=a, \\
(x+y-z)(-x+y+z)=b, \\
(-x+y+z)(x-y+z)=c .
\end{array}\right.
$$

If \( a b c > 0 \), multiply the above three equations and take the square root, we get
$$
(x-y+z)(x+y-z)(-x+y+z)= \pm \sqrt{a b c}
$$

Divide (1) by each of the above equations, the two sets of solutions for the system of equations are:
$$
\begin{array}{l}
x= \pm \frac{b+c}{2 b c} \sqrt{a b c} \text {, } \\
y= \pm \frac{a+c}{2 a c} \sqrt{a b c} \text {, } \\
z= \pm \frac{a+b}{2 a b} \sqrt{a b c} \text {. } \\
\end{array}
$$

The signs in the fractions should be taken as all positive or all negative.
If \( a b c < 0 \) or \( a b c = 0 \), similar discussions can be made.