Mechanics in the contraction $e^{ijk}e_{ijk}$ $e^{ijk}$ is the permutation symbol of $1,2,3$ assuming values $1$ and $-1$ depending on the sign of the permutation. Otherwise the value is zero.
The result is $6,$ but I am not following the mechanics of the operation. I look at it as $e^{123}e_{123}=1 \times 1,$ or $e^{213}e_{213}=-1 \times -1.$
But I guess I should be able to contract one index at a time, as in $e^{1jk}e_{1jk}$ or $e^{2jk}e_{2jk}$ and then proceed with the other two symbols. So in the case of $e^{1jk}e_{1jk}$ if $j=1$ then I end up with $e^{11k}=e_{11k}=0.$ So $j$ must be either $2$ or $3.$ In either case the value of $k$ is determined. So there are really $3! =3\times 2 \times 1$ possible permutations that produce $1$ or $-1.$
Is it correct then to assume that $e^{ijk}$ and $e_{ijk}$ when dealing with valid permutations always yield the same result, i.e. either $1$ or $-1,$ which get multiplied, yielding always $1$?
Why even have upper and lower indices then?
 A: I'm not sure your context of this symbol. (maybe from general relativity?)
The $\epsilon$ symbol itself is not a tensor and does not subject to the same rule as tensors under basis change. Thus in terms of value, the two versions are the same thing. Just written differently to follow the rule that upper and lower index contracts:
By convention, the upper index and lower index version $\epsilon$ are used to contract the vectors of the counter index (lower index $\epsilon_{ijk}$ for upper index vectors $v^i,v^j,v^k$. The upper and lower index vectors (contravariant vector and covarint vector) have a different rule of change under basis change.
For the first question, you are right in that $e^{ijk}e_{ijk}=6$. See wiki for LeviCivita symbol
Because $i,j,k$ each have 3 values, it's a summation over 27 possibilities. But the total number of $ijk$ with no duplicating index is the total permutation number of 3 $3!$.
Then since $e^{ijk}$ and $e_{ijk}$ have the same sign, you just sum up $1$ $6$ times.

More details about contraction. Contraction is just an abbreviation of summation over all the values of each index. So in the full version, we shall write
$$
e^{ijk}e_{ijk}=\sum_{i=1}^{3}\sum_{j=1}^{3}\sum_{k=1}^{3}e^{ijk}e_{ijk}\\
=\sum_{i=1}^{3}\sum_{j=1}^{3}\sum_{k=1}^{3} (e^{ijk})^2\\
=\sum_{(i,j,k)\in S(1,2,3)}(e^{ijk})^2\\
=\sum_{(i,j,k)\in S(1,2,3)} 1\\
=|S(1,2,3)| = 3!
$$
$S(1,2,3)$ is all the permutations of the set ${1,2,3}$
