$$ \begin{aligned} & a=\hat{i}+\alpha \hat{j}+3 \hat{k} \\ & b=3 \hat{i}+\alpha \hat{j}+\hat{k} \end{aligned} $$ area of parallelogram $=\left|\begin{array}{ll}\mathrm{W} & \mathrm{W} \\ \mathrm{b}\end{array}\right|=8 \sqrt{3}$. $$ \begin{aligned} & \underset{a \times b}{\otimes}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 3 \\ 3 & -\alpha & 1 \end{array}\right|=\hat{i}(4 \alpha)-\hat{j}(-8)+\hat{k}(-4 \alpha) \\ & \therefore|\vec{a} \times \vec{b}|=\sqrt{64+32 \alpha^2}=8 \sqrt{3} \\ & \Rightarrow 2+\alpha^2=6 \Rightarrow a^2=4 \\ & \therefore a \cdot b=3-\alpha^2+3=2 \end{aligned} $$