$$\begin{aligned} & =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x \\ & =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}}|\sin x-\cos x| d x \\ & =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{3}}(\sin x-\cos x) d x \\ & =-1+2 \sqrt{2}-\sqrt{3} \\ & =\alpha+\beta \sqrt{2}+\gamma \sqrt{3} \\ & \alpha=-1, \beta=2, \gamma=-1 \\ & 3 \alpha+4 \beta-\gamma=6 \end{aligned}$$