= Solution
Let $D=\mathbf a\cdot(\mathbf b\times\mathbf c)\ne0$ be the <scalar triple product>. The <reciprocal basis> satisfies $\mathbf a\cdot\mathbf a'=1$ and all mismatched dot products vanish, with cyclic analogues. Apply the given <vector triple product> identity to obtain
$$
\mathbf b'\times\mathbf c'
=\frac{(\mathbf c\times\mathbf a)\times(\mathbf a\times\mathbf b)}{D^2}
=\frac{D\mathbf a}{D^2}=\frac{\mathbf a}{D}.
$$
Dotting with $\mathbf a'$ gives
$$
\boxed{\mathbf a'\cdot(\mathbf b'\times\mathbf c')=\frac1D.}
$$
The sign is retained: a negatively oriented original <basis> has a negatively oriented <reciprocal basis>. Equivalently, if the original vectors are columns of $B$, their reciprocals are columns of $B^{-T}$, whose <determinant> is $1/\det B$.
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