= Solution
Let $\Delta=\mathbf a\cdot(\mathbf b\times\mathbf c)\ne0$. 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> are
$$
\boxed{\lambda_1=\frac{\mathbf r\cdot(\mathbf b\times\mathbf c)}{\Delta},\qquad
\mu_1=\frac{\mathbf r\cdot\mathbf a}{\Delta}.}
$$
The <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.
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