I am to prove $\displaystyle \ det[g_{ij}]=\ V^2$ where $\displaystyle \ g_{ij}$ is metric tensor and $\displaystyle \ V=\epsilon_i\cdot\epsilon_j\times\epsilon_k$

Now,I was wondering if the identity

$\displaystyle \epsilon_{ijk}\epsilon_{lmn}=

\begin{vmatrix}

\ g_{11}&\ g_{21}&\ g_{31}\\

g_{12}&\ g_{22}&\ g_{32}\\

g_{13}&\ g_{23}&\ g_{33}

\end{vmatrix}

$

is correct---in which case the problem is solved easily.Note that the antisymmetric tensors are both co-variant.

Surely,the best way to know whether it is correct or not is to prove it...but somehow I am not getting it.If it is correct can anyone give me some hint?