If $$\vec{a} = \alpha\hat{i} + \beta\hat{j} + 3\hat{k}$$, $$\vec{b} = -\beta\hat{i} - \alpha\hat{j} - \hat{k}$$ and $$\vec{c} = \hat{i} - 2\hat{j} - \hat{k}$$ such that $$\vec{a} \cdot \vec{b} = 1$$ and $$\vec{b} \cdot \vec{c} = -3$$, then $$\frac{1}{3}\left((\vec{a} \times \vec{b}) \cdot \vec{c}\right)$$ is equal to ________.

We have $$\vec{a} = \alpha\hat{i} + \beta\hat{j} + 3\hat{k}$$, $$\vec{b} = -\beta\hat{i} - \alpha\hat{j} - \hat{k}$$, $$\vec{c} = \hat{i} - 2\hat{j} - \hat{k}$$, with $$\vec{a} \cdot \vec{b} = 1$$ and $$\vec{b} \cdot \vec{c} = -3$$.

$$\vec{a} \cdot \vec{b} = -\alpha\beta - \alpha\beta - 3 = -2\alpha\beta - 3 = 1$$, so $$\alpha\beta = -2$$ $$-(1)$$.

$$\vec{b} \cdot \vec{c} = -\beta(1) + (-\alpha)(-2) + (-1)(-1) = -\beta + 2\alpha + 1 = -3$$, so $$2\alpha - \beta = -4$$ $$-(2)$$.

From $$(2)$$: $$\beta = 2\alpha + 4$$. Substituting into $$(1)$$: $$\alpha(2\alpha+4) = -2$$, giving $$2\alpha^2 + 4\alpha + 2 = 0$$, i.e., $$(\alpha+1)^2 = 0$$, so $$\alpha = -1$$.

Then $$\beta = 2(-1) + 4 = 2$$. So $$\vec{a} = -\hat{i} + 2\hat{j} + 3\hat{k}$$ and $$\vec{b} = -2\hat{i} + \hat{j} - \hat{k}$$.

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 3 \\ -2 & 1 & -1 \end{vmatrix} = \hat{i}(-2-3) - \hat{j}(1+6) + \hat{k}(-1+4) = -5\hat{i} - 7\hat{j} + 3\hat{k}$$.

$$(\vec{a} \times \vec{b}) \cdot \vec{c} = (-5)(1) + (-7)(-2) + (3)(-1) = -5 + 14 - 3 = 6$$.

$$\frac{1}{3}((\vec{a} \times \vec{b}) \cdot \vec{c}) = \frac{6}{3} = 2$$.

The answer is 2.