If $$\sin \left(\frac{2A + B}{2}\right) = \cos \left(\frac{2A - B}{2}\right) = \frac{\sqrt{3}}{2}, 0^\circ < \frac{2A + B}{2} < 90^\circ$$ and $$0^\circ < \frac{2A + B}{2} < 90^\circ$$ then find the value of $$\sin[3(A - B)]$$.

A

$$\frac{\sqrt{3}}{2}$$

B

$$\frac{1}{\sqrt{2}}$$

C

$$\frac{1}{2}$$

D

1