$A,B,C$ are angles of triangle.

$-x+y\cos C+z\cos B=0$

$x\cos C-y+z\cos A=0$

$x\cos B+y\cos A-z=0$

$\left|A\right|=\left|\begin{array}{ccc}-1& \cos C& \cos B\\ \cos C& -1& \cos A\\ \cos B& \cos A& -1\end{array}\right|$

$\left|A\right|=-1\left(1-{\cos }^{2}A\right)-\cos C\left(-\cos C-\cos A.\cos B\right)+\cos B\left(\cos A.\cos C+\cos B\right)$

$\left|A\right|=-1+{\cos }^{2}A+{\cos }^{2}C+\cos A.\cos B.\cos C+\cos A.\cos B.\cos C+{\cos }^{2}B$

$\because A,B, C$ are angles of triangle.

So, $\left|A\right|=0$ for equilateral triangle and $\left|A\right|\ne 0$ for all other triangles.

Hence, a non-zero solution if $△ABC$ is an equilateral triangle and not for all triangle.