If the area of the parallelogram with \(\bar{a}\) and \(\bar{b}\) as two adjacent sides is 15 square units, then the area (in square units) of the parallelogram, having \(3 \bar{a}+2 \bar{b}\) and \(\bar{a}+3 \bar{b}\) as two adjacent sides, is

Area of parallelogram is \(|\bar{a} \times \bar{b}|\)
\(\therefore \quad|\bar{a} \times \bar{b}|=15\)
\(\therefore \quad\) Area of the required parallelogram
\(\begin{aligned}
& =|(3 \overline{\mathrm{a}}+2 \overline{\mathrm{~b}}) \times(\overline{\mathrm{a}}+3 \overline{\mathrm{~b}})| \\
& =|3(\overline{\mathrm{a}} \times \overline{\mathrm{a}})+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{~b}} \times \overline{\mathrm{a}})+6(\overline{\mathrm{~b}} \times \overline{\mathrm{b}})| \\
& =0+9|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|-2|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|+0 \\
& =7|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| \\
& =7 \times 15 \\
& =105 \text { sq. units }
\end{aligned}\)