If $$0 \leq x < \frac{\pi}{2}$$, then the number of values of $$x$$ for which $$\sin x - \sin 2x + \sin 3x = 0$$, is:

We need to find the number of values of $$x$$ in $$[0, \frac{\pi}{2})$$ satisfying $$\sin x - \sin 2x + \sin 3x = 0$$.

We use the sum-to-product identity. Group the first and third terms:

$$\sin x + \sin 3x = 2 \sin\left(\frac{x + 3x}{2}\right) \cos\left(\frac{3x - x}{2}\right) = 2 \sin 2x \cos x$$

Substituting back into the equation:

$$2 \sin 2x \cos x - \sin 2x = 0$$

Factoring out $$\sin 2x$$:

$$\sin 2x (2\cos x - 1) = 0$$

This gives us two cases:

Case 1: $$\sin 2x = 0$$

$$2x = n\pi$$ where $$n$$ is an integer, so $$x = \frac{n\pi}{2}$$.

In the interval $$[0, \frac{\pi}{2})$$: $$x = 0$$ (when $$n = 0$$) is valid. Note that $$x = \frac{\pi}{2}$$ (when $$n = 1$$) is NOT included since the domain is strictly less than $$\frac{\pi}{2}$$.

Case 2: $$2\cos x - 1 = 0$$

$$\cos x = \frac{1}{2}$$

$$x = \frac{\pi}{3}$$

Since $$\frac{\pi}{3} \approx 1.047$$ and $$\frac{\pi}{2} \approx 1.571$$, we have $$0 < \frac{\pi}{3} < \frac{\pi}{2}$$, so $$x = \frac{\pi}{3}$$ is valid.

Therefore, the valid solutions in $$[0, \frac{\pi}{2})$$ are $$x = 0$$ and $$x = \frac{\pi}{3}$$.

The number of values of $$x$$ is $$2$$.

The correct answer is Option C: 2.