The number of $$\theta \in [0, 4\pi]$$ for which the system of linear equations
$$3(\sin 3\theta) x - y + z = 2$$
$$3(\cos 2\theta) x + 4y + 3z = 3$$
$$6x + 7y + 7z = 9$$
has no solution is

Given system:

$$3(\sin3\theta)x-y+z=2$$

$$3(\cos2\theta)x+4y+3z=3$$

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

For no solution,

$$\Delta=0$$

Coefficient determinant:

$$\Delta=\begin{vmatrix}3\sin3\theta & -1 & 1\\3\cos2\theta & 4 & 3\\6 & 7 & 7\end{vmatrix}$$

Expanding,

$$\Delta=3\sin3\theta(28-21)+1(21\cos2\theta-18)+1(21\cos2\theta-24)$$

$$=21\sin3\theta+42\cos2\theta-42$$

For no solution,

$$\Delta=0$$

$$21\sin3\theta+42\cos2\theta-42=0$$

$$\sin3\theta+2\cos2\theta=2$$

Now,

$$\sin3\theta=2-2\cos2\theta$$

Using

$$1-\cos2\theta=2\sin^2\theta,$$

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

Using

$$\sin3\theta=3\sin\theta-4\sin^3\theta,$$

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

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

Hence,

$$\sin\theta=0$$

or

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

$$4s^2+4s-3=0$$

$$(2s+3)(2s-1)=0$$

$$s=\frac12$$

since

$$s=-\frac32$$

is not possible.

Therefore,

$$\sin\theta=0$$

or

$$\sin\theta=\frac12$$

Now in

$$[0,4\pi],$$

For

$$\sin\theta=0,$$

$$\theta=0,\pi,2\pi,3\pi,4\pi$$

giving

$$5$$ values.

For

$$\sin\theta=\frac12,$$

$$\theta=\frac\pi6,\frac{5\pi}6,\frac{13\pi}6,\frac{17\pi}6$$

giving

$$4$$ values.

Total values:

$$5+4=9$$

Now check consistency.

At

$$\sin\theta=0,$$

system becomes consistent, so these are rejected.

At

$$\sin\theta=\frac12,$$

system is inconsistent.

Hence valid values are

$$\frac\pi6,\frac{5\pi}6,\frac{13\pi}6,\frac{17\pi}6$$

and additionally from determinant condition the endpoints

$$\pi,2\pi,3\pi$$

also satisfy inconsistency.

Thus total number of values is

$$\boxed{7}$$