One side and one diagonal of a parallelogram are represented by \(3 \hat{i}+\hat{j}-\hat{k}\) and \(2 \hat{i}+\hat{j}-2 \hat{k}\) respectively, then the area of parallelogram in square units is

\(\overline{\mathrm{a}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}} \text { and } \overline{\mathrm{c}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
In \(\triangle \mathrm{ABC}\),
\(\begin{aligned}
& \overline{\mathrm{a}+\overline{\mathrm{b}}}=\overline{\overline{\mathrm{c}}} \ldots \ldots[\text { Using triangle law of addition] } \\
& \begin{aligned}
\Rightarrow \overline{\mathrm{b}} & =\overline{\mathrm{c}}-\overline{\mathrm{a}} \\
& =2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}-(3 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=-\hat{\mathrm{i}}-\hat{\mathrm{k}} \\
\overline{\mathrm{a}} \times \overline{\mathrm{b}} & =\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
3 & 1 & -1 \\
-1 & 0 & -1
\end{array}\right|=-\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}
\end{aligned}
\end{aligned}\)
\(\therefore \text { Area of parallelogram } =|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| \)
\( =\sqrt{1+16+1} \)
\( =\sqrt{18}=3 \sqrt{2} \text { sq. units}\)