Let $\vec a,\vec b$ and $\vec c$ be three unit vectors such that $\vec a×(\vec b×\vec c)=\frac{\sqrt{3}}{2} (\vec b+\vec c)$. If $\vec b$ is not parallel to, then the angle between $\vec a$ and $\vec b$ is

$\frac{5π}{6}$

We have,

$\vec a×(\vec b×\vec c)=\frac{\sqrt{3}}{2} (\vec b+\vec c)$

$⇒(\vec a.\vec c)\vec b-(\vec a.\vec b)\vec c=\frac{\sqrt{3}}{2} (\vec b+\vec c)$

$⇒\left(\vec a.\vec c-\frac{\sqrt{3}}{2}\right)\vec b-\left(\vec a.\vec b+\frac{\sqrt{3}}{2}\right)\vec c=\vec 0$

$⇒\vec a.\vec c-\frac{\sqrt{3}}{2}=0$ and $\vec a.\vec b+\frac{\sqrt{3}}{2}=0$  [$∵\vec b$ and $\vec c$ are not parallel vectors]

$⇒\cos θ_1=\frac{\sqrt{3}}{2}$ and $\cos θ_2=-\frac{\sqrt{3}}{2}$, where $θ_1$ and $θ_2$ are angles made by $\vec a$ with $\vec c$ and $\vec b$ respectively.

$⇒θ_1=\frac{π}{6}$ and $θ_2=\frac{5π}{6}$