 Solution

Let $\Delta =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$. Taking the scalar product with $\mathbf{b}\times \mathbf{c}$ in the first expansion kills its other two terms. Taking the scalar product with $\mathbf{a}$ in the second expansion does the same. Thus the two requested coordinates areThe scalar triple product condition guarantees linear independence and the existence of these reciprocal bases. The second coordinate is not obtained by merely normalizing the first basis vector: the two expansions use different bases.