Explanation

(b)
Let angle between $\overrightarrow{\mathbf{b}}\ and\overrightarrow{\mathbf{c}}$ is α.
Given, $\left| \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}} \right| = \sqrt{15}$
$\Longrightarrow \left| \overrightarrow{\mathbf{b}} \right|\left| \overrightarrow{\mathbf{c}} \right|\sin{\alpha = \sqrt{15}}$
$\Longrightarrow \sin\alpha = \frac{\sqrt{15}}{4}$
$\therefore\cos{\alpha = \sqrt{1 - \sin^{2}\alpha} = \sqrt{1 - \frac{15}{16}}}$
$= \frac{1}{4}$
$\because\ \overrightarrow{\mathbf{b}} - 2\overrightarrow{\mathbf{c}}$=λ $\overrightarrow{\mathbf{a}}$ [given]
$\Longrightarrow (\overrightarrow{\mathbf{b}} - 2\overrightarrow{\mathbf{c}})^{2} = \lambda^{2}(\overrightarrow{\mathbf{a}})^{2}$
$\Longrightarrow \ {\overrightarrow{\mathbf{b}}}^{2} + 4{\overrightarrow{\mathbf{c}}}^{2} - 4\overrightarrow{\mathbf{b}} \bullet \overrightarrow{\mathbf{c}} = \lambda^{2}{\overrightarrow{\mathbf{a}}}^{2}$
$\Longrightarrow 16 + 4 \times 1 - 4(|\overrightarrow{\mathbf{b}}|\ |\overrightarrow{\mathbf{c}}|\ cos\ \alpha) = \lambda^{2} \bullet 1^{2}$
$\Longrightarrow 20 - 4 = \lambda^{2}$
$\Longrightarrow \ \lambda = \pm 4$