If three vectors $\vec a,\vec b,\vec c$ are such that $\vec a≠0$ and $\vec a× \vec b = 2(\vec a×\vec c),|\vec a|=|\vec c|=1,|\vec b|=4$ and the angle between $\vec b$ and $\vec c$ is $\cos^{-1}(1/4)$, then $\vec b-2\vec c=λ\vec a$ where λ is equal to:

-4

$(\vec a× \vec b) = 2(\vec a×\vec c)⇒\vec a×(\vec b-2\vec c)=0⇒\vec b-2\vec c=λ\vec a$   ....(i)

Also, $\vec b.\vec c=|\vec b||\vec c|cosθ=1×4×\frac{1}{4}=1$

Squaring (i), we get: 

$|\vec b|^2-4|\vec c|^2-4\vec b.\vec c=λ^2|\vec a|^2⇒16+4-4=λ^2⇒λ=±4$