For unit vectors, \(|\vec{a} - \vec{b}|^2 = 2 - 2\vec{a} \cdot \vec{b}\), and similarly for the others. So

Sum = \(6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})\).

Since \(|\vec{a} + \vec{b} + \vec{c}|^2 = 3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \ge 0\), we get \(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \ge -\frac{3}{2}\).

Sum ≤ \(6 + 3 = 9\). The maximum 9 is reached when \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\).

Answer: (B) 9