Explanation

(b)
Given that, $\left| \overrightarrow{\mathbf{a}} \right| = 3,\ |\overrightarrow{\mathbf{b}}| = 4$ and $\overrightarrow{\mathbf{a}} + \lambda\overrightarrow{\mathbf{b}}$ is perpendicular to $\overrightarrow{\mathbf{a}} - \lambda\overrightarrow{\mathbf{b}}$.
$\therefore\left( \mathbf{\ }\overrightarrow{\mathbf{a}}\mathbf{+}\lambda\overrightarrow{\mathbf{b}} \right)\mathbf{\bullet}\left( \overrightarrow{\mathbf{a}}\mathbf{-}\lambda\overrightarrow{\mathbf{b}}\mathbf{\ } \right) = 0$
$\Rightarrow \overrightarrow{\mathbf{a}}\mathbf{\bullet}\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}}\lambda + \lambda\overrightarrow{\mathbf{b}}\mathbf{\bullet}\overrightarrow{\mathbf{a}} - \lambda^{2}\overrightarrow{\mathbf{b}}\mathbf{\bullet}\overrightarrow{\mathbf{b}} = 0$
$\Rightarrow \left| \overrightarrow{\mathbf{a}} \right|^{2} - \lambda^{2}\left| \overrightarrow{\mathbf{b}} \right|^{2} = 0\$
$\Rightarrow \ \lambda^{2} = \frac{\left| \overrightarrow{\mathbf{a}} \right|^{2}}{\left| \overrightarrow{\mathbf{b}} \right|^{2}}\ \Rightarrow \ \lambda = \frac{\left| \overrightarrow{\mathbf{a}} \right|}{\left| \overrightarrow{\mathbf{b}} \right|} = \frac{3}{4}$