Solution

Length of the two diagonals will be
$$\Rightarrow$$ $$d_{1}=|(5a+2b)+(a-3b)|$$
$$\Rightarrow$$ $$d_{2}=|(5a+2b)-(a-3b)|$$

$$\Rightarrow$$ $$d_{1}=|6a-b|$$ and $$\Rightarrow$$ $$d_{2}=|4a+5b|$$
So,
$$d_{1}=\sqrt{|6a|^2+|-b|^2+2|6a| |-b|cos(\pi-\frac{\pi}{4})}$$
$$d_1=\sqrt{36(2\sqrt{2})^2+3^2+12(2\sqrt2)3(-\frac{1}{\sqrt2})}$$
$$d_1=15$$

$$d_{2}=\sqrt{|4a|^2+|5b|^2+2|4a| |5b|cos(\frac{\pi}{4})}$$
$$d_2=\sqrt{16(2\sqrt{2})^2+5\times 3^2+40\times 2\sqrt2\times 3\times {\frac{1}{\sqrt2}}}$$
$$d_2=\sqrt{593}>15$$

$$\therefore\ d_{2}>d_{1}$$

Hence, length of longer diagonal is $$\sqrt{593}$$.