have a solution $$\left( {{x_0},{y_0},{z_0}} \right)$$ with $${y_0}{z_0}{\mkern 1mu} \ne {\mkern 1mu} 0,$$ is
View the equation in xyz, y and t.
We have,
$$(xyz)\sin 3\theta - y\cos 3\theta - z\cos 3\theta = 0$$
$$(xyz)\sin 3\theta - 2y\sin 3\theta - 2z\cos 3\theta = 0$$
$$(xyz)\sin 3\theta - y(\cos 3\theta + \sin 3\theta ) - 2z\cos 3\theta = 0$$
$$xyz \ne 0$$
Hence, the equation has non-trivial solution which gives
$$\left| {\matrix{ {\sin 3\theta } & { - \cos 3\theta } & { - \cos 3\theta } \cr {\sin 3\theta } & { - 2\sin 3\theta } & { - 2\cos 3\theta } \cr {\sin 3\theta } & { - (\cos 3\theta + \sin 3\theta )} & { - 2\cos 3\theta } \cr } } \right| = 0$$
$$ \Rightarrow \sin 3\theta \cos 3\theta (\sin 3\theta - \cos 3\theta ) = 0$$
$$ \Rightarrow \sin 3\theta = 0$$ then $$xyz = 0$$ (not possible)
$$\cos 3\theta = 0$$ not possible
$$\sin 3\theta = \cos 3\theta \Rightarrow \tan 3\theta = 1$$
$$3\theta = n\pi + {\pi \over 4},n \in z$$
$$\theta = {{n\pi } \over 3} + {\pi \over {12}}$$ ; $$\theta = {\pi \over {12}},{{5\pi } \over {12}},{{9\pi } \over {12}}$$
Thus there are 3 solutions.