We can see that $-1\leq x\leq1$, so we can let $x=\sin\alpha$. Same with $y$ and $z$: let $y=\sin\beta$ and $z=\sin\gamma$. WLOG we can let $-\frac{\pi}{2}\leq\alpha, \beta, \gamma\leq\frac{\pi}{2}$. This makes $\sqrt{1-x^2}=\cos\alpha$ and the same with the other two. The equations now are $a\sin\alpha+b\sin\beta=c\sin\gamma$ $a\cos\alpha+b\cos\beta=c\cos\gamma$. We square both equations and add them to get $c^2=a^2+b^2+2ab(\sin\alpha\sin\beta+\cos\alpha\cos\beta)=a^2+b^2+2ab\cos(\alpha-\beta)$. Since $-\frac{\pi}{2}\leq\alpha, \beta\leq\frac{\pi}{2}$, then $\cos(\alpha-\beta)$ can take any value from -1 to 1. So $-2ab\leq c^2-a^2-b^2\leq 2ab\Leftrightarrow$ $a^2-2ab+b^2=(a-b)^2\leq c^2\leq a^2+2ab+b^2=(a+b)^2\Leftrightarrow$ $|a-b|\leq c\leq a+b$. This is a condition, and we need to show that it is the only condition. So if we let $\cos^{-1}\left(\frac{c^2-a^2-b^2}{2ab}\right)=\theta$, then we need to find when $\alpha-\beta=\theta$ or $\alpha-\beta=-\theta$. We can let $\beta\in[-\frac{\pi}{2}, \frac{\pi}{2}-\theta]$ and let $\alpha=\beta+\theta$, and we can also switch $\beta$ and $\alpha$. So once we've found $\alpha$ and thus $\beta$, we can let $\gamma=\sin^{-1}\left(\frac{a\sin\alpha+b\sin\beta}{c}\right)$ and the formation of $\alpha$ and $\beta$ guarantees that $a\cos\alpha+b\cos\beta=c\cos\gamma$. In summary, the only conditions on $a, b, c$ are that $|a-b|\leq c\leq a+b$, and all solutions are of the form $(x, y, z)=(\sin\alpha, \sin\beta, \sin\gamma)$ where $|\alpha-\beta|=\cos^{-1}\left(\frac{c^2-a^2-b^2}{2ab}\right)$ and $\gamma=\sin^{-1}\left(\frac{a\sin\alpha+b\sin\beta}{c}\right)$.