Given,

$$3(\sin 3\theta )x - y + z = 2$$

$$3(\cos 2\theta )x + 4y + 3z = 3$$

$$6x + 7y + 7z = 9$$

For no solutions determinant of coefficient will be = 0

$$\therefore$$ $$D = \left| {\matrix{ {3\sin 3\theta } & { - 1} & 1 \cr {3\cos 2\theta } & 4 & 3 \cr 6 & 7 & 7 \cr } } \right| = 0$$

$$ \Rightarrow 3\sin 3\theta (28 - 21) + 1(21\cos 2\theta - 18) + 1(21\cos 2\theta - 24) = 0$$

$$ \Rightarrow 21\sin 3\theta + 42\cos 2\theta - 42 = 0$$

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

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

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

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

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

$$\therefore$$ $$\sin \theta = 0$$

$$ \Rightarrow \theta = \pi ,2\pi ,3\pi $$ when $$\theta \in (0,4\pi )$$

or,

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

$$ \Rightarrow 4{\sin ^2}\theta + 6\sin \theta - 2\sin \theta - 3 = 0$$

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

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

$$\therefore$$ $$\sin \theta = {1 \over 2}$$

or,

$$\sin \theta = - {3 \over 2}$$ [not possible as $$\sin \in [ - 1,1]$$]

$$\therefore$$ $$\sin \theta = {1 \over 2}$$

$$ \Rightarrow \theta = {\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}$$

$$\therefore$$ Possible values of $$\theta = \pi ,2\pi ,3\pi ,{\pi \over 6},{{5\pi } \over 6},{{13\pi } \over 6},{{17\pi } \over 6}$$

$$\therefore$$ Total 7 values of $$\theta$$ possible.