Let $$\vec{a} = \hat{i} + \alpha\hat{j} + 3\hat{k}$$ and $$\vec{b} = 3\hat{i} - \alpha\hat{j} + \hat{k}$$. If the area of the parallelogram whose adjacent sides are represented by the vectors $$\vec{a}$$ and $$\vec{b}$$ is $$8\sqrt{3}$$ square units, then $$\vec{a} \cdot \vec{b}$$ is equal to ______.

We have $$\vec{a} = \hat{i} + \alpha\hat{j} + 3\hat{k}$$ and $$\vec{b} = 3\hat{i} - \alpha\hat{j} + \hat{k}$$. The area of the parallelogram is $$|\vec{a} \times \vec{b}| = 8\sqrt{3}$$.

Computing the cross product: $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 3 \\ 3 & -\alpha & 1 \end{vmatrix} = \hat{i}(\alpha \cdot 1 - 3 \cdot (-\alpha)) - \hat{j}(1 \cdot 1 - 3 \cdot 3) + \hat{k}(1 \cdot (-\alpha) - \alpha \cdot 3) = (4\alpha)\hat{i} + 8\hat{j} + (-4\alpha)\hat{k}$$.

The magnitude is $$|\vec{a} \times \vec{b}| = \sqrt{16\alpha^2 + 64 + 16\alpha^2} = \sqrt{32\alpha^2 + 64}$$.

Setting this equal to $$8\sqrt{3}$$: $$32\alpha^2 + 64 = 192$$, so $$32\alpha^2 = 128$$, giving $$\alpha^2 = 4$$.

Now, $$\vec{a} \cdot \vec{b} = (1)(3) + (\alpha)(-\alpha) + (3)(1) = 3 - \alpha^2 + 3 = 6 - \alpha^2 = 6 - 4 = 2$$.

Therefore, $$\vec{a} \cdot \vec{b} = 2$$.