The length of longer diagonal of the parallelogram constructed on   $5\overline{\mathrm{a}}+2\overline{\mathrm{b}}$ and  $\overline{\mathrm{a}}-3\overline{\mathrm{b}}$  as sides, if it is given $\left|\overline{\mathrm{a}}\right|=2\sqrt{2} ,\left|\overline{\mathrm{b}}\right|=3$ and $(\overline{\mathrm{a}},\overline{\mathrm{b}})=\mathrm{\pi }/4$ is

$\begin{array}{l}\overline{{\mathrm{d}}_1}=5\overline{\mathrm{a}}+2\overline{\mathrm{b}}+\overline{\mathrm{a}}-3\overline{\mathrm{b}}=6\overline{\mathrm{a}}-\overline{\mathrm{b}} ;\overline{{\mathrm{d}}_2}=(5\overline{\mathrm{a}}+2\overline{\mathrm{b}})-(\overline{\mathrm{a}}-3\overline{\mathrm{b}})=4\overline{\mathrm{a}}+5\overline{\mathrm{b}}\\ \left|\overline{\mathrm{a}}\right|=2\sqrt{2},\left|\overline{\mathrm{b}}\right|=3,(\overline{\mathrm{a}},\overline{\mathrm{b}})=\mathrm{\pi }/4\Rightarrow \overline{\mathrm{a}}\cdot \overline{\mathrm{b}}=\left|\overline{\mathrm{a}}\right|\left|\overline{\mathrm{b}}\right|\cos \left(\overline{\mathrm{a}},\overline{\mathrm{b}}\right)=2\sqrt{2}\times 3\times \frac{1}{\sqrt{2}}=6\\ \overline{{\mathrm{d}}_1}=|6\overline{\mathrm{a}}-\overline{\mathrm{b}}|=\sqrt{36|\overline{\mathrm{a}}{|}^2+|\overline{\mathrm{b}}{|}^2-12(\overline{\mathrm{a}}\cdot \overline{\mathrm{b}})} \Rightarrow \sqrt{36\left(8\right)+9-12\left(6\right)}=\sqrt{225}=15\text{ units }\\ \left|\overline{{\mathrm{d}}_2}\right|=|4\overline{\mathrm{a}}+5\overline{\mathrm{b}}|=\sqrt{16|\overline{\mathrm{a}}{|}^2+25|\overline{\mathrm{b}}{|}^2+40(\overline{\mathrm{a}}\cdot \overline{\mathrm{b}})}\Rightarrow \sqrt{16\left(8\right)+25\left(9\right)+40\left(6\right)}=\sqrt{593}\end{array}$
Length of longer diagonal is $\sqrt{593}$