It is important to understand that usually the determinant is a scalar
but in this case it is not. Recall the cross product $A \times B$,
can be computed as a determinant and it is a vector. We will interpret
as determinant to the regular expansion of determinants but this time
instead of a scalar entries we have 2-rank tensor entries.
We expand the determinant as follows:
\begin{eqnarray*}
\left |
\begin{array}{ccc}
\delta_m^r & \delta_n^r & \delta_p^r \\
\delta_m^s & \delta_n^s & \delta_p^s \\
\delta_m^t & \delta_n^t & \delta_p^t
\end{array}
\right |
&=& \delta_m^r \delta_n^s \delta_p^t + \delta_n^r \delta_p^s \delta_m^t +
\delta_p^r \delta_m^s \delta_n^t
- \delta_p^r \delta_n^s \delta_m^t - \delta_n^r \delta_m^s \delta_p^t - \delta_m^r \delta_p^s
\delta_n^t .
\end{eqnarray*}
We define
\begin{eqnarray*}
a_{mnp}^{rst} =
&=&
\delta_m^r \delta_n^s \delta_p^t +
\delta_m^r \delta_n^s \delta_p^t + \delta_m^r \delta_n^s \delta_p^t -
\delta_m^r \delta_n^s \delta_p^t - \delta_m^r \delta_n^s \delta_p^t
-\delta_m^r \delta_n^s \delta_p^t
\end{eqnarray*}
where we commuted factors so $m,n,p$ are in that order in all subsscripts.
We recognize that the sequence of delta operations $a_{mnp}^{rst}$ is a completely skew-symmetri
system in $r,s,t$. That is, for example, let us reverse the order or $s$ and $t$ and find
\begin{eqnarray*}
a^{rts}_{mnp} &=&
\delta_m^r \delta_n^t \delta_p^s +
\delta_m^s \delta_n^r \delta_p^t + \delta_m^t \delta_n^s \delta_p^r -
\delta_m^s \delta_n^t \delta_p^r - \delta_m^t \delta_n^r \delta_p^s -\delta_m^r \delta_n^s
\delta_p^t \\
&=& -\delta_m^r \delta_n^s \delta_p^t -\delta_m^t \delta_n^r \delta_p^s -\delta_m^s \delta_n^t
\delta_p^r + \delta_m^t \delta_n^s \delta_p^r + \delta_m^s \delta_n^r \delta_p^t + \delta_m^r
\delta_n^t \delta_p^s \\
&=& -a^{rst}_{mnp}
\end{eqnarray*}
where we fixed the order by using the commutative law of the sum (with sign) operation .
Likewise we can interchange the order of $r,s$ and $r,t$ an get
\begin{eqnarray*}
a_{mnp}^{srt} = -a_{mnp}^{rst} \quad \text{and} \quad
a_{mnp}^{tsr} = -a_{mnp}^{rst}.
\end{eqnarray*}
Hence $a_{mnp}^{rst}$ ($m,n,p$ fixed) is completely skew-symmetric in $r,s,t$ and so
\begin{eqnarray*}
a_{mnp}^{rst} = \epsilon^{rst} a_{mnp}^{123}
\end{eqnarray*}
with
\begin{eqnarray*}
a_{mnp}^{123} =
\delta_m^1 \delta_n^2 \delta_p^3 +
\delta_m^3 \delta_n^1 \delta_p^2 + \delta_m^2 \delta_n^3 \delta_p^1 -
\delta_m^3 \delta_n^2 \delta_p^1 - \delta_m^2 \delta_n^1 \delta_p^3 -\delta_m^1 \delta_n^3
\delta_p^2 .
\end{eqnarray*}
This is equal to $1$ if $mpn$ is an even permutation of $(1 \; 2 \; 3)$ or
$-1$ if $mpn$ is an odd permutation of $(1 \; 2 \; 3)$. That is,
\begin{eqnarray*}
a_{mnp}^{rst} = \epsilon^{rst} \epsilon_{mnp}
\end{eqnarray*}
if $m=n$ or $m=p$ we would get 0. Then
\begin{eqnarray*}
\left |
\begin{array}{ccc}
\delta_m^r & \delta_n^r & \delta_p^r \\
\delta_m^s & \delta_n^s & \delta_p^s \\
\delta_m^t & \delta_n^t & \delta_p^t
\end{array}
\right |
&=& \epsilon^{rst} \epsilon_{mnp} .
\end{eqnarray*}
