If $a+b+c\ne 0$the system of equations $\left(b+c\right)\left(y+z\right)-ax=b-c,$$\left(c+a\right)\left(z+x\right)-by=c-a, \left(a+b\right)\left(x+y\right)-cz=a-b$  has

$\left(b+c\right)\left(y+z\right)-ax=b-c\_\_\_\_\_\_\_\left(1\right)$

$\left(a+b\right)\left(x+y\right)-cz=a-b\_\_\_\_\_\_\_\left(2\right)$

$\left(c+a\right)\left(z+x\right)-by=c-a\_\_\_\_\_\_\_\left(3\right)$

From equations  1,2, and 3 we have

$x=\frac{c-b}{a+b+c}, y=\frac{a-c}{a+b+c}, z=\frac{b-a}{a+b+c}$