Kronecker delta and Levi-Civita symbol Prove that
$$
det
        \left[
        \begin{matrix}
        δ_{ak} & δ_{al} & δ_{am} \\
        δ_{bk} & δ_{bl} & δ_{bm} \\
        δ_{ck} & δ_{cl} & δ_{cm} \\
        \end{matrix}
\right]
=ε_{abc}ε_{klm}
$$
with $a,b,c,k,l,m \in \{1,2,3\}$ where $δ_{ij}$ is Kronecker delta and $ε_{ijk}$ is Levi-Civita symbol in dimension 3
I found this problem in a very old mathematical-competition.I would apreciate if we can find a solution for this.

Write the Kronecker Delta as
$$\delta_{ij}=\hat x_i \cdot \hat x_j \tag 1$$
in terms of then inner product of Cartesian unit vectors.  Write the Levi-Civita symbol as
$$\epsilon_{ijk}=\hat x_i\cdot(\hat x_j \times \hat x_k) \tag 2$$
in terms of the scalar triple product of Cartesian unit vectors.  Using $(2)$, we have
$$\begin{align}
\epsilon_{abc}\epsilon_{k\ell m}&=\left(\hat x_a\cdot(\hat x_b \times \hat x_c)\right)\left(\hat x_k\cdot(\hat x_{\ell} \times \hat x_m)\right)\tag3\\\\
&=\delta_{ak}\left((\hat x_b \times \hat x_c)\cdot \hat x_k\hat x_k\cdot(\hat x_{\ell} \times \hat x_m)\right)\tag4\\\\
&+\delta_{a\ell}\left((\hat x_b \times \hat x_c)\cdot \hat x_{\ell}\hat x_{\ell}\cdot(\hat x_m\times \hat x_k)\right)\\\\
&+\delta_{am}\left((\hat x_b \times \hat x_c)\cdot \hat x_{m}\hat x_{m}\cdot(\hat x_k\times \hat x_{\ell})\right)\\\\
&=\delta_{ak}\left((\hat x_b \times \hat x_c)\cdot (\hat x_{\ell} \times \hat x_m)\right)\\\\
&+\delta_{a\ell}\left((\hat x_b \times \hat x_c)\cdot (\hat x_m\times \hat x_k)\right)\\\\
&+\delta_{am}\left((\hat x_b \times \hat x_c)\cdot (\hat x_k\times \hat x_{\ell})\right)\\\\
&=\delta_{ak}\left(\delta_{b\ell}\delta_{cm}-\delta_{bm}\delta_{c\ell}\right)\\\\
&+\delta_{a\ell}\left(\delta_{bm}\delta_{ck}-\delta_{bk}\delta_{cm}\right)\\\\
&+\delta_{am}\left(\delta_{bk}\delta_{c\ell}-\delta_{b\ell}\delta_{ck}\right)\\\\
&=
det
        \left[
        \begin{matrix}
        δ_{ak} & δ_{al} & δ_{am} \\
        δ_{bk} & δ_{bl} & δ_{bm} \\
        δ_{ck} & δ_{cl} & δ_{cm} \\
        \end{matrix}
\right]
\end{align}$$
as was to be shown!

NOTE: In going from $(3)$ to $(4)$, we used the result in This Answer.
