If $$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\sqrt{1 - \sin 2x} \, dx = \alpha + \beta\sqrt{2} + \gamma\sqrt{3}$$, where $$\alpha, \beta$$ and $$\gamma$$ are rational numbers, then $$3\alpha + 4\beta - \gamma$$ is equal to ______.

We need to evaluate $$\int_{\pi/6}^{\pi/3} \sqrt{1 - \sin 2x} \, dx$$.

Simplify the integrand: using the identity $$1 - \sin 2x = \sin^2 x + \cos^2 x - 2\sin x \cos x = (\sin x - \cos x)^2$$:

$$\sqrt{1 - \sin 2x} = |\sin x - \cos x|$$

Determine the sign on the interval: $$\sin x = \cos x$$ when $$x = \frac{\pi}{4}$$.

For $$x \in [\pi/6, \pi/4]$$: $$\cos x > \sin x$$, so $$|\sin x - \cos x| = \cos x - \sin x$$.

For $$x \in [\pi/4, \pi/3]$$: $$\sin x > \cos x$$, so $$|\sin x - \cos x| = \sin x - \cos x$$.

Evaluate the integral: $$ I = \int_{\pi/6}^{\pi/4} (\cos x - \sin x) \, dx + \int_{\pi/4}^{\pi/3} (\sin x - \cos x) \, dx $$

$$ I = [\sin x + \cos x]_{\pi/6}^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{\pi/3} $$

First part: $$\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right) = \sqrt{2} - \frac{1 + \sqrt{3}}{2}$$

Second part: $$\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = -\frac{1 + \sqrt{3}}{2} + \sqrt{2}$$

$$ I = \sqrt{2} - \frac{1 + \sqrt{3}}{2} + \sqrt{2} - \frac{1 + \sqrt{3}}{2} = 2\sqrt{2} - (1 + \sqrt{3}) $$

$$ I = -1 + 2\sqrt{2} - \sqrt{3} $$

Match with the given form: $$I = \alpha + \beta\sqrt{2} + \gamma\sqrt{3}$$ where $$\alpha = -1, \beta = 2, \gamma = -1$$.

$$ 3\alpha + 4\beta - \gamma = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6 $$

The answer is $$\boxed{6}$$.