If area of the parallelogram with \(\bar{a}\) and \(\bar{b}\) as two adjacent sides is 20 square units, then the area of the parallelogram having \(3 \bar{a}\) \(+\overline{\mathrm{b}}\) and \(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}\) as two adjacent sides in square units is

We have \(|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|=20\) and we have to find value of
\(
\begin{aligned}
& |(3 \overline{\mathrm{a}}+\overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}})| \\
& (3 \overline{\mathrm{a}}+\overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+3 \overline{\mathrm{b}}) \\
& =6(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+2(\overline{\mathrm{b}} \times \overline{\mathrm{a}})+3(\overline{\mathrm{b}} \times \overline{\mathrm{b}}) \\
& =0+9(\overline{\mathrm{a}} \times \overline{\mathrm{b}})-2(\overline{\mathrm{a}} \times \overline{\mathrm{b}})+0=7(\overline{\mathrm{a}} \times \overline{\mathrm{b}})
\end{aligned}
\)
Hence area of the required parallelogram \(=7 \times 20=140\)