$\overrightarrow{ b }=3 \hat{ i }-\alpha \hat{ j }+\hat{ k }$
area of parallelogram $=|\overrightarrow{ a } \times \overrightarrow{ b }|=8 \sqrt{3}$.
$\overrightarrow{ a } \times \overrightarrow{ b }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 1 & \alpha & 3 \\ 3 & -\alpha & 1\end{array}\right|=\hat{ i }(4 \alpha)-\hat{ j }(-8)+\hat{ k }(-4 \alpha)$
$\therefore|\overrightarrow{ a } \times \overrightarrow{ b }|=\sqrt{64+32 \alpha^{2}}=8 \sqrt{3}$
$\Rightarrow 2+\alpha^{2}=6 \Rightarrow \alpha^{2}=4$
$\therefore \overrightarrow{ a } \cdot \overrightarrow{ b }=3-\alpha^{2}+3=2$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.