Let \(\overline{\mathrm{a}}=3 \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\overline{\mathrm{b}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\). If the area of the parallelogram whose adjacent sides are represented by the vectors \(\bar{a}\) and \(\bar{b}\), is \(8 \sqrt{3}\) sq. units, then \(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}\) is equal to

\(\begin{aligned}
\overline{\mathrm{a}} \times \overline{\mathrm{b}} & =\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
3 & -\alpha & 1 \\
1 & \alpha & 3
\end{array}\right| \\
& =-4 \alpha \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+4 \alpha \hat{\mathrm{k}}
\end{aligned}\)
Area of parallelogram \(=|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|\)
\(\begin{aligned}
& \Rightarrow 8 \sqrt{3}=\sqrt{16 \alpha^2+64+16 \alpha^2} \\
& \Rightarrow 8 \sqrt{3}=\sqrt{32 \alpha^2+64}
\end{aligned}\)
Squaring on both sides, we get
\(\begin{aligned}
& 192=32 \alpha^2+64 \\
& \Rightarrow 32 \alpha^2=128 \\
& \Rightarrow \alpha^2=4 \\
& \begin{aligned}
& \overline{\mathrm{a}} \cdot \overline{\mathrm{~b}}=(3 \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\
& \quad=3-\alpha^2+3 \\
& \quad=6-4 \\
& \quad=2
\end{aligned}
\end{aligned}\)