Let a, b and c be three vectors satisfying a × b=(a ×...

Let $\theta$ be the angle between $\mathbf{b} and$\mathbf{c}$. Given,$|\mathrm{b} \times \mathrm{c}|=\sqrt{15}$\begin{array}{l} \Rightarrow|\mathrm{b}\| \mathrm{c}| \sin \theta=\sqrt{15}\\ \Rightarrow \sin \theta=\frac{\sqrt{15}}{|\mathrm{~b}| \mathrm{c} \mid}\\ \Rightarrow \sin \theta=\sqrt{\frac{15}{4 \times 1}}=\frac{\sqrt{15}}{4}\\ \quad \therefore \cos \theta=\sqrt{1-\frac{15}{16}}=\frac{1}{\sqrt{16}}=\frac{1}{4} \end{array}$$Now given,$\mathrm{b}-2 \mathrm{c}=\lambda \mathrm{a} \Rightarrow\left|\mathrm{b}-2 \mathrmc^{2}\right|=\mid \lambda \mathrma^{2}$\Rightarrow|$\mathrm{b}|^{2}+$4|$\mathrm{c}|-$4($\mathrm{~b} \cdot \mathrm{c})=\lambda^{2}|\mathrm{a}|^{2} $$\begin{array}{l} \Rightarrow16+4-4|\mathrm{~b}\| \mathrm{c}| \cos \theta=\lambda^{2}\\ \Rightarrow 20-16 \cos \theta=\lambda^{2}\\ \Rightarrow 20-4=\lambda^{2} \Rightarrow \lambda^{2}=16\left(\therefore \cos \theta=\frac{1}{4}\right)\\ \Rightarrow \lambda=\pm4 \end{array}$$