Let $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$ and $$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$$ be two vectors such that $$|\vec{a}| = 1$$; $$\vec{a} \cdot \vec{b} = 2$$ and $$|\vec{b}| = 4$$. If $$\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$$, then the angle between $$\vec{b}$$ and $$\vec{c}$$ is equal to :

Given $$|\vec{a}| = 1$$, $$\vec{a} \cdot \vec{b} = 2$$, $$|\vec{b}| = 4$$, and $$\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$$.

$$\vec{b} \cdot \vec{c} = 2\vec{b} \cdot (\vec{a} \times \vec{b}) - 3|\vec{b}|^2 = 0 - 3(16) = -48$$

(since $$\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$$)

$$|\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 - 12(\vec{a} \times \vec{b}) \cdot \vec{b} + 9|\vec{b}|^2$$

$$= 4|\vec{a} \times \vec{b}|^2 + 144$$

$$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 = 16 - 4 = 12$$.

$$|\vec{c}|^2 = 48 + 144 = 192$$

$$\cos\theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|} = \frac{-48}{4\sqrt{192}} = \frac{-48}{4 \cdot 8\sqrt{3}} = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = \frac{-\sqrt{3}}{2}$$

$$\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = 150°$$

The answer is Option (3): $$\boxed{\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)}$$.