If \(|\bar{a}|=2,|\bar{b}|=3\) and \(\bar{a}, \bar{b}\) are mutually perpendicular vectors, then the area of the triangle whose vertices are \(0, a+2 b, a-2 b\) is

Let position vectors of \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) be \(0, a+2 b, a-2 b\)
\(\begin{aligned}
& \text {Area of } \triangle \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \\
& =\frac{1}{2}|(\overline{\mathrm{a}}+2 \overline{\mathrm{~b}})(\overline{\mathrm{a}}-2 \overline{\mathrm{~b}})| \\
& =\frac{1}{2}|\overline{\mathrm{a}} \times \overline{\mathrm{a}}-\overline{\mathrm{a}} \times 2 \overline{\mathrm{~b}}+2 \overline{\mathrm{~b}} \times \overline{\mathrm{a}}=2 \overline{\mathrm{~b}} \times \overline{\mathrm{b}}| \\
& =\frac{1}{2}|2 \overline{\mathrm{~b}} \times \overline{\mathrm{a}}+2 \overline{\mathrm{~b}} \times \overline{\mathrm{a}}| \\
& =\frac{1}{2} \times 4|\overline{\mathrm{~b}} \times \overline{\mathrm{a}}| \\
& =2 \times 2 \times 3 \\
& =12 \text { sq. units. }
\end{aligned}\)