Let $A=\sin 3\theta,\qquad B=\cos 3\theta.$ Then the given system becomes

1. $(y+z)B=(xyz)A$
2. $xA=\frac{2B}{y}+\frac{2A}{z}$
3. $(xyz)A=(y+2z)B+yA$

We are asked for the number of values of $\theta$ with $0<\theta<\pi$ for which this system has a solution $(x_0,y_0,z_0)$ with $y_0z_0\neq 0$.

1. Compare equations (1) and (3)

From (1) and (3), since both are equal to $(xyz)A$, we get $(y+z)B=(y+2z)B+yA.$ So $yB+zB=yB+2zB+yA$ $zB=2zB+yA$ $yA+zB=0.$ Hence $y\sin 3\theta+z\cos 3\theta=0.$ So $yA=-zB.$ Since $y,z\neq 0$, this relation is important.

2. Use this in equation (1)

Equation (1) is $(y+z)B=xyzA.$ Using $yA=-zB$, if $A\neq 0$, then $y=-\frac{zB}{A}.$ Substitute into (1): $\left(-\frac{zB}{A}+z\right)B=\left(x\cdot -\frac{zB}{A}\cdot z\right)A.$ Simplify: $z\left(1-\frac{B}{A}\right)B=-xz^2B.$ This is messy, so instead we proceed more cleanly by solving via ratio.

From $yA=-zB,$ we get $y=-\frac{B}{A}z \qquad (A\neq 0).$ Then $y+z=z\left(1-\frac{B}{A}\right)=z\frac{A-B}{A}.$ Substitute into (1): $z\frac{A-B}{A}B=xyzA.$ Now $yz=-\frac{B}{A}z^2,$ so RHS is $x\left(-\frac{B}{A}z^2\right)A=-xBz^2.$ Thus $z\frac{(A-B)B}{A}=-xBz^2.$ If $B\neq 0$ and $z\neq 0$,

so $xz=\frac{B-A}{A}.$

3. Use equation (2)

Equation (2): $xA=\frac{2B}{y}+\frac{2A}{z}.$ Using $y=-\frac{B}{A}z$, $\frac{2B}{y}=\frac{2B}{-(B/A)z}=-\frac{2A}{z}.$ Hence $\frac{2B}{y}+\frac{2A}{z}= -\frac{2A}{z}+\frac{2A}{z}=0.$ Therefore $xA=0.$ So either

1. $A=0$, or
2. $x=0$.

If $A\neq 0$, then $x=0$. But from above, $xz=\frac{B-A}{A}.$ Since $z\neq 0$, this gives $0=\frac{B-A}{A} \implies B=A.$ Thus for $A\neq 0$, a necessary condition is $\sin 3\theta=\cos 3\theta.$ So $\tan 3\theta=1.$

4. Check special cases $A=0$ or $B=0$

We must separately examine them because we divided by $A$ and $B$.

Case 1: $A=0$

Then $\sin 3\theta=0$. Equations become:

1. $(y+z)B=0$
2. $0=\frac{2B}{y}$
3. $0=(y+2z)B$ Since $y\neq 0$, equation (2) implies $B=0.$ But $A=0$ and $B=0$ cannot happen simultaneously. So no solution.

Case 2: $B=0$

Then $\cos 3\theta=0$. Equations become:

1. $0=xyzA$
2. $xA=\frac{2A}{z}$
3. $xyzA=yA$ Since $B=0$, we have $A=\pm 1\neq 0$. Then from (1), $xyz=0.$ Because $y,z\neq 0$, this gives $x=0$. But then from (2), $0=\frac{2A}{z},$ impossible since $A\neq 0$ and $z\neq 0$. So no solution.

Hence only the condition $A=B$ can work.

5. Solve $\sin 3\theta=\cos 3\theta$

This gives $\tan 3\theta=1$ $3\theta=\frac{\pi}{4}+n\pi.$ Therefore $\theta=\frac{\pi}{12}+\frac{n\pi}{3}.$ Now use $0<\theta<\pi$: $0<\frac{\pi}{12}+\frac{n\pi}{3}<\pi.$ Multiplying by $12/\pi$, $0<1+4n<12.$ So $-1<4n<11.$ Thus integer values are $n=0,1,2.$ Hence the possible values are $\theta=\frac{\pi}{12},\ \frac{5\pi}{12},\ \frac{3\pi}{4}.$ So the number of possible values is $3$.

6. Verify existence of a solution for these values

When $A=B\neq 0$, equation $yA+zB=0$ becomes $y+z=0 \implies y=-z.$ Then equation (1): $(y+z)B=xyzA \implies 0=xyzA.$ Since $A\neq 0$ and $y,z\neq 0$, we get $x=0.$ Equation (2): $0=\frac{2B}{y}+\frac{2A}{z}=2A\left(\frac{1}{y}+\frac{1}{z}\right).$ With $y=-z$, this holds. Equation (3): $0=(y+2z)B+yA=(2y+2z)A=2A(y+z)=0.$ So solutions indeed exist, e.g. $(x,y,z)=(0,1,-1).$ Thus all such $\theta$ are valid.

Final answer

The number of possible values of $\theta$ is $\boxed{3}.$