Step-by-step Solution:

Given:

\[ \vec{e_1} = (1, 1, 1), \quad \vec{e_2} = (1, 1, -1) \]

Let \( \vec{a} \) and \( \vec{b} \) be two vectors such that:

\[ \vec{e_1} = 2\vec{a} + \vec{b} \quad \text{and} \quad \vec{e_2} = \vec{a} + 2\vec{b} \]

From the given equations:

\[ \vec{e_1} = (1, 1, 1) \implies 2\vec{a} + \vec{b} = (1, 1, 1) \]

\[ \vec{e_2} = (1, 1, -1) \implies \vec{a} + 2\vec{b} = (1, 1, -1) \]

From the first equation:

\[ \vec{b} = (1, 1, 1) - 2\vec{a} \]

Substituting this into the second equation:

\[ \vec{a} + 2((1, 1, 1) - 2\vec{a}) = (1, 1, -1) \]

\[ \vec{a} + (2, 2, 2) - 4\vec{a} = (1, 1, -1) \]

\[ -3\vec{a} + (2, 2, 2) = (1, 1, -1) \]

\[ -3\vec{a} = (-1, -1, -3) \implies \vec{a} = \left(\frac{1}{3}, \frac{1}{3}, 1\right) \]

Now substitute \( \vec{a} \) back into \( \vec{b} = (1, 1, 1) - 2\vec{a} \):

\[ \vec{b} = (1, 1, 1) - 2\left(\frac{1}{3}, \frac{1}{3}, 1\right) = \left(\frac{1}{3}, \frac{1}{3}, -1\right) \]

The angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \) is given by:

\[ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\| \|\vec{b}\|} \]

First, calculate \( \vec{a} \cdot \vec{b} \):

\[ \vec{a} \cdot \vec{b} = \left(\frac{1}{3}\right)\left(\frac{1}{3}\right) + \left(\frac{1}{3}\right)\left(\frac{1}{3}\right) + (1)(-1) = \frac{1}{9} + \frac{1}{9} - 1 = -\frac{7}{9} \]

Next, calculate \( \|\vec{a}\| \) and \( \|\vec{b}\| \):

\[ \|\vec{a}\| = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + (1)^2} = \sqrt{\frac{1}{9} + \frac{1}{9} + 1} = \sqrt{\frac{11}{9}} = \frac{\sqrt{11}}{3} \]

\[ \|\vec{b}\| = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + (-1)^2} = \sqrt{\frac{1}{9} + \frac{1}{9} + 1} = \sqrt{\frac{11}{9}} = \frac{\sqrt{11}}{3} \]

Finally, calculate \( \cos\theta \):

\[ \cos\theta = \frac{-\frac{7}{9}}{\frac{\sqrt{11}}{3} \cdot \frac{\sqrt{11}}{3}} = \frac{-\frac{7}{9}}{\frac{11}{9}} = -\frac{7}{11} \]

Therefore, the angle between \( \vec{a} \) and \( \vec{b} \) is:

\[ \theta = \cos^{-1}\left(-\frac{7}{11}\right) \]