Since $ \mathbf{b} \times \mathbf{c} $ is orthogonal to both $ \mathbf{b} $ and $ \mathbf{c}, $ $ \mathbf{a} \cdot \mathbf{b} = 0 $ and $ \mathbf{a} \cdot \mathbf{c} = 0 $. Since $ \|\mathbf{a} + \mathbf{b} + \mathbf{c}\| = 1, $ \[(\mathbf{a} + \mathbf{b} + \mathbf{c}) \cdot (\mathbf{a} + \mathbf{b} + \mathbf{c}) = 1.\]Expanding, we get \[\mathbf{a} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c} + 2 \mathbf{a} \cdot \mathbf{b} + 2 \mathbf{a} \cdot \mathbf{c} + 2 \mathbf{b} \cdot \mathbf{c} = 1.\]From what we know, this becomes \[\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + 0 + 0 + 2 \mathbf{b} \cdot \mathbf{c} = 1.\]Hence, $ \mathbf{b} \cdot \mathbf{c} = 0 $. This means the angle between $ \mathbf{b} $ and $ \mathbf{c} $ is $ \boxed{90^\circ} $.