$\overline{a}$ and $\overline{\mathrm{c}}$ are unit vectors and $\left|\stackrel{\mathrm{‾}}{\mathrm{b}}\right|=4$ with $\stackrel{\mathrm{‾}}{\mathrm{a}}\times \stackrel{\mathrm{‾}}{\mathrm{b}}=2\stackrel{\mathrm{‾}}{\mathrm{a}}\times \stackrel{\mathrm{‾}}{\mathrm{c}}$. The angle between  $\overline{b} and \overline{c}$  is ${\cos }^{-1} \left(\frac{1}{4}\right)$. Then $\stackrel{\mathrm{‾}}{\mathrm{b}}-2\stackrel{\mathrm{‾}}{\mathrm{c}}=\mathrm{\lambda }\stackrel{\mathrm{‾}}{\mathrm{a}}$ , if $\lambda$ is

Given, $\left|\stackrel{\mathrm{‾}}{\mathrm{a}}\right|=\left|\stackrel{\mathrm{‾}}{\mathrm{c}}\right|=1$ and

 $$\begin{gathered} Given a\times b=2a\times c \\ \Rightarrow a\times b-2a\times c=0 \\ \Rightarrow a\times (b-2c)==0 \\ a and (b-2c) are collinear \\ \Rightarrow b-2c=\lambda a \\ \Rightarrow {\left|b-2c\right|}^2={\lambda }^2{\left|a\right|}^2 \\ \Rightarrow {\left|b\right|}^2+4{\left|c\right|}^2-4\overline{b}.\overline{c}={\lambda }^2{\left|a\right|}^2 \\ \Rightarrow 16+4-4.4.1.\frac{1}{4}={\lambda }^2.\left(1\right) \\ \Rightarrow {\lambda }^2=16 \end{gathered}$$