= Solution
Write the row components as $E_{ai}=(e_a)_i$. Expanding the <cross product> in components,
$$
(e_1\times e_2)\cdot e_3=\epsilon_{ijk}(e_1)_i(e_2)_j(e_3)_k
$$
equals the alternating six-term expansion of the <determinant> of $E$. Therefore \b[$(e_1\times e_2)\cdot e_3=\det E$].
Geometrically, $|e_1\times e_2|$ is the area of the parallelogram spanned by $e_1,e_2$, and its direction is perpendicular to that plane. A nonzero <scalar triple product> means both that this area is nonzero and that $e_3$ has a nonzero component normal to the plane. Thus the vectors are <linearly independent> and form a <basis> of $\mathbb R^3$, without needing to be orthogonal. Equivalently, their parallelepiped has nonzero volume.
The entries of $E\hat E^T$ are $e_a\cdot\hat e_b$, so the required duality is exactly $E\hat E^T=I$. Since $E$ is invertible, this has the unique solution
$$
\boxed{\hat E^T=E^{-1},\qquad \hat E=E^{-T}}.
$$
Its <determinant> is $1/\det E\ne0$, so its rows are again a <basis>. Put $\Delta=(e_1\times e_2)\cdot e_3$. The <cross product> $e_2\times e_3$ is perpendicular to $e_2,e_3$, and its <dot product> with $e_1$ is $\Delta$. Consequently the <reciprocal basis> is
$$
\boxed{\hat e_1=\frac{e_2\times e_3}{\Delta},\qquad \hat e_2=\frac{e_3\times e_1}{\Delta},\qquad \hat e_3=\frac{e_1\times e_2}{\Delta}}.
$$
These vectors have precisely the required <dot products>, so uniqueness identifies them with the rows of $E^{-T}$.
For the final request, an <orthonormal basis> of rows gives $EE^T=I$. If instead the reciprocal rows are <orthonormal>, then $\hat E\hat E^T=I$ and substitution of $\hat E=E^{-T}$ gives $E^{-T}E^{-1}=I$, again implying $EE^T=I$. Thus in either case \b[$E$ is an <orthogonal matrix>] and $\hat E=E$.
The further printed claim that $E$ must be a <rotation matrix> is false without an orientation assumption. The <matrix> $E=\operatorname{diag}(1,1,-1)$ has orthonormal rows and is its own <reciprocal basis> <matrix>, but $\det E=-1$: it is a <reflection>. By the <orientation of an orthonormal reciprocal basis>, the corrected conclusion is
$$
\boxed{E\in O(3),\qquad E\text{ is a rotation matrix if and only if }\det E=+1.}
$$
A right-handed original ordered <basis> supplies exactly this missing condition.
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