If $$3 \tan \theta = 2\sqrt{3} \sin \theta, 0^\circ < \theta < 90^\circ$$, then find the value of $$2 \sin^2 2\theta - 3 \cos^2 3\theta$$.

$$3\tan\theta=2\sqrt{3}\sin\theta$$

$$3\frac{\sin\theta\ }{\cos\theta\ }=2\sqrt{3}\sin\theta$$

$$\cos\theta\ =\frac{3}{2\sqrt{3}}$$

$$\cos\theta\ =\frac{\sqrt{3}}{2}$$

$$\theta\ =30^{\circ\ }$$ [$$0^\circ < \theta < 90^\circ$$]

$$2\sin^22\theta-3\cos^23\theta=2\sin^260^{\circ\ }-3\cos^290^{\circ\ }$$

= $$2\left(\frac{\sqrt{3}}{2}\right)^2-3\left(0\right)^2$$

= $$\frac{3}{2}$$

Hence, the correct answer is Option B