4・129 Solve the system of equations $\left\{\begin{array}{l}x^{2}=6+(y-z)^{2}, \\ y^{2}=2+(z-x)^{2}, \\ z^{2}=3+(x-y)^{2} .\end{array}\right.$
[Solution] Rewrite the original system of equations as
$$
\left\{\begin{array}{l}
6=x^{2}-(y-z)^{2}=(x-y+z)(x+y-z), \\
2=y^{2}-(z-x)^{2}=(y-z+x)(y+z-x), \\
3=z^{2}-(x-y)^{2}=(z-x+y)(z+x-y) .
\end{array}\right.
$$
(1) $\times$ (2) $\times$ (3) gives
$$
36=(x-y+z)^{2}(y-z+x)^{2}(z-x+y)^{2} \text {, }
$$

Thus, $\quad(x-y+z)(y-z+x)(z-x+y)= \pm 6$.
(5) $\div$ (1) gives $z-x+y= \pm 1$.
(5) $\div$ (2) gives $x-y+z= \pm 3$.
(5) $\div$ (3) gives $y-z+x= \pm 2$.
(6) + (7) + (8) gives $x+y+z= \pm 6$.
(9) - (6) gives $2 x= \pm 5, x= \pm \frac{5}{2}$,
(9) - (7) gives $2 y= \pm 3, y= \pm \frac{3}{2}$,
(9) - (8) gives $2 z= \pm 4, z= \pm 2$.
Therefore, the original system of equations has two sets of solutions
$$
\begin{array}{l}
x_{1}=\frac{5}{2}, y_{1}=\frac{3}{2}, z_{1}=2 \\
x_{2}=\frac{-5}{2}, y_{2}=\frac{-3}{2}, z_{2}=-2
\end{array}
$$