If $$x^2 + 4y^2 + 3z^2 + \frac{19}{4} = 2\sqrt{3}(x + y + z)$$, then the value of $$(x - 4y + 3z)$$ is:

A

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

B

$$2\sqrt{3}$$

C

$$\sqrt{3}$$

D

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