Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$\vec{a} = \sqrt{14}$$, $$\vec{b} = \sqrt{6}$$ and $$\vec{a} \times \vec{b} = \sqrt{48}$$. Then $$(\vec{a} \cdot \vec{b})^2$$ is equal to ______.

Applying the Lagrange identity for vectors, we have $$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$$. Substituting the given magnitudes yields $$ (\sqrt{48})^2 + (\vec{a} \cdot \vec{b})^2 = (\sqrt{14})^2 \times (\sqrt{6})^2$$. Simplifying these expressions gives $$48 + (\vec{a} \cdot \vec{b})^2 = 14 \times 6 = 84$$ and therefore $$ (\vec{a} \cdot \vec{b})^2 = 84 - 48 = 36\,. $$