Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}$. 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
$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}$
Area of parallelogram $=|\vec{a} \times \vec{b}|$
$=\mid(\hat{i}+\alpha \hat{j}+3 \hat{k}) \times(3 \hat{i}-\alpha \hat{j}+\hat{k})$
$8 \sqrt{3}=|(4 \alpha) \hat{i}+8 \hat{j}-(4 \alpha) \hat{k}|$
$(64)(3)=16 \alpha^{2}+64+16 \alpha^{2}$
$(64)(3)=32 \alpha^{2}+64$
$6=\alpha^{2}+2$
$\alpha^{2}=4$
$\therefore \quad \vec{a}=\hat{i}+\alpha \hat{j}+3 \hat{k}$
$\vec{b}=3 \hat{i}-\alpha \hat{j}+\hat{k}$
$\vec{a} \cdot \vec{b}=3-\alpha^{2}+3$
$=6-\alpha^{2}$
$=6-4$
$=2$