If \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}\) are three vectors, \(|\overline{\mathrm{a}}|=2,|\overline{\mathrm{b}}|=4,|\overline{\mathrm{c}}|=1\), \(|\overline{\mathrm{b}} \times \overline{\mathrm{c}}|=\sqrt{15}\) and \(\overline{\mathrm{b}}=2 \overline{\mathrm{c}}+\lambda \overline{\mathrm{a}}\), then the value of \(\lambda\), is

If angle between \(\bar{b}\) and \(\bar{c}\) is \(\alpha\) and
\(
\begin{aligned}
& |\overline{\mathrm{b}} \times \overline{\mathrm{c}}|=\sqrt{15} \\
& \Rightarrow|\overline{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \sin \alpha=\sqrt{15} \\
& \Rightarrow \sin \alpha=\frac{\sqrt{15}}{4} \\
& \Rightarrow \cos \alpha=\frac{1}{4}
\end{aligned}
\)
Now, \(\overline{\mathrm{b}}-2 \overline{\mathrm{c}}=\lambda \overline{\mathrm{a}}\)
\(
\begin{aligned}
& \Rightarrow|\overline{\mathrm{b}}-2 \overline{\mathrm{c}}|^2=\lambda^2|\overline{\mathrm{b}}|^2 \\
& \Rightarrow|\overline{\mathrm{b}}|^2+4|\overline{\mathrm{c}}|^2-4 \overline{\mathrm{b}} \cdot \overline{\mathrm{c}}=\left.\left.\lambda^2\right|^{-\overline{\mathrm{a}}}\right|^2 \\
& \Rightarrow 16+4-4\{|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \cos \alpha\}=\lambda^2 \\
& \Rightarrow 16+4-4 \times 4 \times 1 \times \frac{1}{4}=\lambda^2 \\
& \Rightarrow \lambda^2=16 \\
& \Rightarrow \lambda= \pm 4
\end{aligned}
\)