$x=A(1+a)(1+b)(1+c)$

$y=B(2+a)(2+b)(2+c)$

$z=C(3+a)(3+b)(3+c)$

$\frac{x}{1+a}+\frac{y}{2+a}+\frac{z}{3+a}$

$=A(1+b)(1+c)+B(2+b)(2+c)+C(3+b)(3+c)$

$=A(1+b+c+bc)+B(4+2b+2c+bc)+C(9+3b+3c+bc)$

All that has to equal to $1$, so we get a system of equations:

$A+4B+9C=1$

$A+2B+3C=0$

$A+B+C=0$

Which if we solve gives us: $A=0.5$, $B=-1$, $C=0.5$.

By symmetry it's easy to show that this solutions works for all three of the above equations.