The area (in sq. units) of the parallelogram whose diagonals are along the vectors \(8 \hat{i}-6 \hat{j}\) and \(3 \hat{i}+4 \hat{j}-12 \hat{k}\), is

Let \(\overline{\mathrm{a}}=8 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}, \overline{\mathrm{b}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}\)
\(\begin{aligned}
& \therefore \quad \text { Area of parallelogram }=\frac{1}{2}|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| \text {. } \\
& \overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
8 & -6 & 0 \\
3 & 4 & -12
\end{array}\right|=72 \hat{\mathrm{i}}+96 \hat{\mathrm{j}}-50 \hat{\mathrm{k}} \\
& \therefore \quad \frac{1}{2}|\overrightarrow{\mathrm{a}} \times \overline{\mathrm{b}}|=\sqrt{5184+9216+2500} \times \frac{1}{2} \\
& =\sqrt{16900} \times \frac{1}{2}=130 \times \frac{1}{2}=65
\end{aligned}\)