The area of the parallelogram whose diagonals are represented by the vectors \(\bar{a}=3 \hat{i}-\hat{j}-2 \hat{k}\) and \(\bar{b}=-\hat{i}+3 \hat{j}-3 \hat{k}\) is

Let \(\overrightarrow{\mathrm{d}_1}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\) and \(\overrightarrow{\mathrm{d}_2}=-\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}\)
\(\therefore\left|\overrightarrow{\mathrm{d}_1}\right|=\sqrt{14}\) and \(\left|\overrightarrow{\mathrm{d}_2}\right|=\sqrt{19}\)
\(
\begin{aligned}
& \text { Also } \overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
3 & -1 & -2 \\
-1 & 3 & -3
\end{array}\right|=9 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+8 \hat{\mathrm{k}} \\
& \therefore\left|\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right|=\sqrt{81+121+64}=\sqrt{266} \\
& \left|\mathrm{~d}_1 \times \mathrm{d}_2\right|^2=\left|\mathrm{d}_1\right|^2\left|\mathrm{~d}_2\right|^2 \sin ^2 \theta \\
& \therefore 266=(14)(19) \sin ^2 \theta \\
& \therefore \sin ^2 \theta=1 \Rightarrow \sin \theta=1 \ldots\left[\because 0 < \theta \leq \frac{\pi}{2}\right]
\end{aligned}
\)
Area of parallelogram \(=\frac{1}{2}\left|\mathrm{~d}_1\right|\left|\mathrm{d}_2\right| \sin \theta\)
\(
=\frac{\sqrt{266}}{2}
\)