If $\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=2\mathrm{\pi },$ then the system of equations

           $\begin{array}{r}\mathrm{x}+(\cos \mathrm{\gamma })\mathrm{y}+(\cos \mathrm{\beta })\mathrm{z}=0\\ (\cos \mathrm{\gamma })\mathrm{x}+\mathrm{y}+(\cos \mathrm{\alpha })\mathrm{z}=0\\ (\cos \mathrm{\beta })\mathrm{x}+(\cos \mathrm{\alpha })\mathrm{y}+\mathrm{z}=0\end{array}$

has

Given $\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=2\mathrm{\pi }$

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{ccc}1& \cos \mathrm{\gamma }& \cos \mathrm{\beta }\\ \cos \mathrm{\gamma }& 1& \cos \mathrm{\alpha }\\ \cos \mathrm{\beta }& \cos \mathrm{\alpha }& 1\end{array}\right|\\ =1-{\cos }^2\mathrm{\alpha }-\cos \mathrm{\gamma }(\cos \mathrm{\gamma }-\cos \mathrm{\alpha cos}\mathrm{\beta })+\cos \beta \left(\cos \alpha \cos \gamma -\cos \beta \right)\\ =1-{\cos }^2\mathrm{\alpha }-{\cos }^2\mathrm{\beta }-{\cos }^2\mathrm{\gamma }+2\cos \mathrm{\alpha cos}\mathrm{\beta cos}\mathrm{\gamma }\\ ={\sin }^2\mathrm{\alpha }-{\cos }^2\mathrm{\beta }-\cos \mathrm{\gamma }(\cos \mathrm{\gamma }-2\cos \mathrm{\alpha cos}\mathrm{\beta })\\ =-\cos (\mathrm{\alpha }+\mathrm{\beta })\cos (\mathrm{\alpha }-\mathrm{\beta })-\cos \mathrm{\gamma }\left(\cos \left(\mathrm{\alpha }+\mathrm{\beta }\right)-2\mathrm{cos\alpha cos\beta }\right)\\ =-\cos (2\mathrm{\pi }-\mathrm{\gamma })\cos (\mathrm{\alpha }-\mathrm{\beta })+\cos \mathrm{\gamma }\left(\cos \left(\mathrm{\alpha }-\mathrm{\beta }\right)\right)=0\end{array}$

So, the system of equation has infinitely many solutions.