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

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

The signs in the fractions are taken as all positive or all negative simultaneously.
If $a b c < 0$ or $a b c = 0$, similar discussions can be made.