Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2$$, $$\vec{a} \cdot \vec{b} = 3$$ and $$|\vec{a} \times \vec{b}|^2 = 75$$. Then $$|\vec{a}|^2$$ is equal to _____

We have $$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2$$, $$\vec{a} \cdot \vec{b} = 3$$, and $$|\vec{a} \times \vec{b}|^2 = 75$$.

Expanding the first condition: $$|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2$$. This simplifies to $$2(\vec{a} \cdot \vec{b}) = |\vec{b}|^2$$, so $$|\vec{b}|^2 = 2 \times 3 = 6$$.

Now we use the identity $$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$$. Substituting: $$75 + 9 = |\vec{a}|^2 \times 6$$, so $$|\vec{a}|^2 = \frac{84}{6} = 14$$.

Hence, the correct answer is $$\boxed{14}$$.