= Solution
A <pseudotensor> has an additional <determinant> factor under an orthogonal transformation. Thus its rank-three transformation law is
$$
\epsilon'_{ijk}=(\det L)L_{ia}L_{jb}L_{kc}\epsilon_{abc}.
$$
The contracted expression $L_{ia}L_{jb}L_{kc}\epsilon_{abc}$ is fully antisymmetric in $i,j,k$, so it equals a scalar multiple of $\epsilon_{ijk}$. For $(i,j,k)=(1,2,3)$ it is precisely the <determinant> expansion of $L$, fixing that scalar to $\det L$. Therefore
$$
\boxed{\epsilon'_{ijk}=(\det L)^2\epsilon_{ijk}=\epsilon_{ijk}.}
$$
The <Levi-Civita symbol> consequently has the same numerical entries in every orthonormal frame when interpreted as a pseudotensor. With an ordinary rank-three <tensor> transformation law alone it would instead acquire a minus sign under an orientation-reversing transformation.
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