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