For a homogeneous system of 3 linear equations in $x,y,z$ to have a non-trivial solution, the determinant of its coefficient matrix must be zero.

The system is: $x+3y+7z=0$ $-x+4y+7z=0$ $\sin 3\theta\, x + \cos 2\theta\, y + 2z=0$

So the coefficient matrix is

Set $\det(A)=0$.

Use the identity $1-\cos2\theta=2\sin^2\theta$ So, $2(1-\cos2\theta)=4\sin^2\theta$

Therefore, $4\sin^2\theta=\sin3\theta$

Expand $\sin3\theta$: $\sin3\theta=3\sin\theta-4\sin^3\theta$

So, $4\sin^2\theta=3\sin\theta-4\sin^3\theta$

Rearranging, $4\sin^3\theta+4\sin^2\theta-3\sin\theta=0$

$\sin\theta\,(4\sin^2\theta+4\sin\theta-3)=0$

Since $\theta\in(0,\pi)$, we have $\sin\theta\neq 0$.

So, $4\sin^2\theta+4\sin\theta-3=0$

Let $s=\sin\theta$. Then $4s^2+4s-3=0$

Solving: $s=\frac{-4\pm\sqrt{16+48}}{8}=\frac{-4\pm 8}{8}$

Thus, $s=\frac{1}{2} \quad \text{or} \quad s=-\frac{3}{2}$

Since $\sin\theta\in[-1,1]$, only $\sin\theta=\frac{1}{2}$ is valid.

In the interval $(0,\pi)$, $\sin\theta=\frac{1}{2}$ gives $\theta=\frac{\pi}{6},\; \frac{5\pi}{6}$

So the number of values of $\theta$ is 2.

Comparing with the options:

• A: two
• B: one
• C: four
• D: three

Hence the correct option is A.