Solution: Compute $ \|\mathbfa + \mathbfb + \mathbfc\|^2 = 3 + 2(\mathbfa \cdot \mathbfb + \mathbfb \cdot \mathbfc + \mathbfc \cdot \mathbfa) = 3 + 2\left(\frac12 + \frac12 + \frac12\right) = 3 + 3 = 6 $. Thus, the norm is $ \sqrt6 $.