Find the value of q satisfying

Given,

$\left|\begin{array}{ccc}1 & 1 & \sin 3 \theta \\ -4 & 3 & \cos 2 \theta \\ 7 & -7 & -2\end{array}\right|=0$

On expanding along $\mathrm{C}_{3}$, we have

$\sin 3 \theta \times(28-21)-\cos 2 \theta \times(-7-7)-2(3+4)=0$

$7 \sin 3 \theta+14 \cos 2 \theta-14=0$

$\sin 3 \theta+2 \cos 2 \theta-2=0$

$\left(3 \sin \theta-4 \sin ^{3} \theta\right)+2\left(1-2 \sin ^{2} \theta\right)-2=0$

$4 \sin ^{3} \theta-4 \sin ^{2} \theta+3 \sin \theta=0$

$\sin \theta\left(4 \sin ^{2} \theta-4 \sin \theta+3\right)=0$

$\sin \theta\left(4 \sin ^{2} \theta-6 \sin \theta+2 \sin \theta+3\right)=0$

$\sin \theta(2 \sin \theta+1)(2 \sin \theta-3)=0$

$\sin \theta=0$ or $\sin \theta \doteq-1 / 2$ or $\sin \theta=3 / 2$

$\theta=n \pi$ or $\theta=m \pi+(-1)^{n}\left(-\frac{\pi}{6}\right) ; m, n \in Z$

$\sin \theta=\frac{-3}{2}$ is not possible