If $\vec{e_1} =(1, 1, 1)$ and $\vec{e_2} = (1,1,-1)$ and $\vec a$ and $\vec b$ are two vectors such that $\vec{e_1} = 2\vec a +\vec b$ and $\vec{e_2} = \vec a +2\vec b$, then angle between $\vec a$ and $\vec b$ is

$\cos^{-1}(-\frac{7}{11})$

We have,

$2\vec a+\vec b=\hat i+\hat j+\hat k$ and $\vec a+2\vec b=\hat i+\hat j-\hat k$

Solving these two equations, we get

$\vec a=\frac{1}{3}(\hat i+\hat j+3\hat k)$ and $\vec b=\frac{1}{3}(\hat i+\hat j-3\hat k)$

$∴\cos θ=\frac{\vec a .\vec b}{|\vec a||\vec b|}⇒\cos θ=-\frac{7}{11}⇒θ=\cos^{-1}(-\frac{7}{11})$