If $$6(\sec^2 59^\circ - \cot^2 31^\circ) + \frac{2}{3}\sin 90^\circ - 3 \tan^2 56^\circ y \tan^2 34^\circ = \frac{y}{3}$$, then the value of $$y$$ is:

A

2

B

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

C

$$\frac{-2}{3}$$

D

-2