The number of all possible values of $\theta$ where $0 < \theta < \pi$, for which the system of equations $(y + z)\cos 3\theta = (xyz)\sin 3\theta$, $x\sin 3\theta = \frac{2\cos 3\theta}{y} + \frac{2\sin 3\theta}{z}$, $(xyz)\sin 3\theta = (y + 2z)\cos 3\theta + y\sin 3\theta$ have a solution $(x_0, y_0, z_0)$ with $y_0 z_0 \neq 0$, is

Given: (y+z)cos 3θ = xyz sin 3θ

$\text{Given: } x\sin 3\theta = \dfrac{2\cos 3\theta}{y} + \dfrac{2\sin 3\theta}{z}$

Given: xyz sin 3θ = (y+2z)cos 3θ + ysin 3θ

$\text{To find: number of } \theta \in (0,\pi)$

$\text{s.t. a solution } (x_0,y_0,z_0) \text{ exists with } y_0 z_0 \neq 0$

$1.\ \text{combine (1), (3)} \Rightarrow \text{relation in } y, z$

$2.\ \text{use (2)} \Rightarrow (y+z)\cos 3\theta = 0$

$3.\ \tan 3\theta = 1, \ \text{count } \theta \in (0,\pi)$

$c = \cos 3\theta, \qquad s = \sin 3\theta$

from (1): xyz s = (y+z)c in (3): xyz s = (y+2z)c + y s

$\Rightarrow (y+z)c = (y+2z)c + y s$

$\Rightarrow -z\,c = y\,s$

$\text{from (2): } x\,s = \dfrac{2c}{y} + \dfrac{2s}{z}$

× yz: xyz s = 2cz + 2sy $xyz\,s = (y+z)c$ $(y+z)c = 2cz + 2sy$

$sy = -zc \ \Rightarrow \ (y+z)c = 2cz - 2cz = 0$

$c = 0 \Rightarrow s = \pm 1$

$-z\,c = y\,s \Rightarrow y\,s = 0 \Rightarrow y = 0 \ \times$

$\text{but } y_0 z_0 \neq 0 \Rightarrow c \neq 0$

$c \neq 0 \Rightarrow y + z = 0 \Rightarrow z = -y$

$-z\,c = y\,s, \quad z = -y \Rightarrow y\,c = y\,s$

$\Rightarrow c = s \Rightarrow \tan 3\theta = 1$

from (1): xyz s = (y+z)c = 0, s ≠ 0 ⇒ xyz = 0; y, z ≠ 0 ⇒ x = 0

$\text{(2): } 0 = \dfrac{2c}{y} + \dfrac{2s}{-y}$

$= \dfrac{2c}{y} - \dfrac{2c}{y} = 0 \ \checkmark$

$\tan 3\theta = 1 \Rightarrow 3\theta = \dfrac{\pi}{4} + n\pi$

$3\theta = \dfrac{\pi}{4}, \ \dfrac{5\pi}{4}, \ \dfrac{9\pi}{4}, \dots$

$\theta = \dfrac{\pi}{12} + \dfrac{n\pi}{3}$

$0 < \theta < \pi: \quad n = 0, 1, 2$

$\theta = \dfrac{\pi}{12}, \ \dfrac{5\pi}{12}, \ \dfrac{3\pi}{4}$

$\boxed{\text{Number of values of } \theta = 3}$