= Solution
Taking the parallelogram spanned by $e_2,e_3$ as base gives base area $|e_2\times e_3|$ and height $|e_1\cdot(e_2\times e_3)|/|e_2\times e_3|$. A tetrahedron has one third of the corresponding pyramid volume and the triangle has half the parallelogram base, hence
$$
V=\frac16|\Delta|,\qquad \Delta=e_1\cdot(e_2\times e_3).
$$
The reciprocal <vectors> are
$$
f_1=\frac{e_2\times e_3}{\Delta},\quad f_2=\frac{e_3\times e_1}{\Delta},\quad f_3=\frac{e_1\times e_2}{\Delta},
$$
which directly satisfy $f_i\cdot e_j=\delta_{ij}$.
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