jmc

Let $p=\frac{x}{a},$ $q=\frac{y}{b},$ $r=\frac{z}{c}.$ Then $p+q+r=3$ and $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0,$ so $pq+pr+qr=0.$

We want $p^2+q^2+r^2.$ Squaring the equation $p+q+r=3,$ we get $$p^2+q^2+r^2+2(pq+pr+qr)=9,$$so $p^2+q^2+r^2=\boxed{9}.$