Explanation

(c)
Given that, $\left| \overrightarrow{\mathbf{a}} \right| = 2\sqrt{2},\left| \overrightarrow{\mathbf{b}} \right| = 3$
The longer vectors is $5\overrightarrow{\mathbf{a}} + 2\overrightarrow{\mathbf{b}} + \ \overrightarrow{\mathbf{a}} - 3\overrightarrow{\mathbf{b}} = 6\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}}$
Length of one diagonal
$= \left| 6\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}} \right|\$
$= \sqrt{36{\overrightarrow{\mathbf{a}}}^{2} + {\overrightarrow{\mathbf{b}}}^{2} - 2 \times 6\left| \overrightarrow{\mathbf{a}} \right|\mathbf{\ }\left| \overrightarrow{\mathbf{b}} \right|\cos{45 \circ}}$
$= \sqrt{36 \times 8 + 9 - 12 \times 2\sqrt{2} \times 3 \times \frac{1}{\sqrt{2}}}$
$= \sqrt{288 + 9 - 12 \times 6} = \sqrt{225} = 15$
Other diagonal is $4\overrightarrow{\mathbf{a}} + 5\overrightarrow{\mathbf{b}}$.
Its length $= \sqrt{16 \times 8 + 25 \times 9 + 40 \times 6} = \sqrt{593}$