Let $\overline{a},\overline{b},\overline{c}$ be three vectors satisfying $\overline{a}\times \overline{b}=2\overline{a}\times \overline{c},$ $\left|\overline{a}\right|=\left|\overline{c}\right|=1,$ $\left|\overline{b}\right|=4$ and $\left|\overline{b}\times \overline{c}\right|=\sqrt{15}.$ If $\overline{b}-2\overline{c}=\lambda \overline{a}$ then  $\lambda$ is ______ 

$$\begin{gathered} |\overline{b}\times \overline{c}{|}^2=15 \\ |\overline{b}{|}^2|\overline{c}{|}^2{\sin }^2\theta =15 \\ 16(1){\sin }^2\theta =15 \\ \Rightarrow {\sin }^2\theta =\frac{15}{16} \end{gathered}$$

$\begin{array}{rcl} & \Rightarrow & {\cos }^2\theta =\frac{1}{16}\\ |\overline{b}-2\overline{c}|& =& \lambda \left|\overline{a}\right|\\ & & squaring on both sides \\ |\overline{b}{|}^2& =& 4|\overline{c}{|}^2-4\overline{b}\cdot \overline{c}\\ & =& {\lambda }^2|\overline{a}{|}^2\end{array}$

$16+4\left(1\right)-4\left(4\right)\left(1\right)\cos \theta ={\lambda }^2\left(1\right)16+4-4\left(4\right)\frac{1}{4}={\lambda }^2$

${\lambda }^2=16\Rightarrow \lambda =\pm 4$