The value of non-zero scalars α and  β such that for all vectors $\vec{a}$  and $\vec{b}$ such that $\alpha (2\vec{a}-\vec{b})+\beta (\vec{a}+2\vec{b})=8\vec{b}-\vec{a}$ is

Given: $\alpha (2\vec a-\vec b)+\beta (\vec a+2\vec b)=8\vec b-\vec a$ for all vectors $\vec a,\vec b$.

Combine like terms: $(2\alpha+\beta)\vec a+(-\alpha+2\beta)\vec b=-\vec a+8\vec b$.

Equate coefficients:
$2\alpha+\beta=-1$,
$-\alpha+2\beta=8$.

Solve:
From $2\alpha+\beta=-1 \Rightarrow \beta=-1-2\alpha$.
Substitute in $-\alpha+2\beta=8$:
$-\alpha+2(-1-2\alpha)=8 \Rightarrow -\alpha-2-4\alpha=8 \Rightarrow -5\alpha=10 \Rightarrow \alpha=-2$.
Then $\beta=-1-2(-2)=3$.

Answer: $\alpha=-2,\ \beta=3$.