Since I've seen you post in some physics sections before, here is an excerpt on how to use the levi-cievta sign from notes i wrote years ago (since physicists love the following two ideas, it's worth getting use to it):
We will now introducte the Levi-Cievta symbol as follows: $$\epsilon_{ijk} = \begin{cases}
1 & 123, 312, 231 \\
-1 & 321, 132, 213 \\
0 & i=j=k
\end{cases} $$
Where the numbers represent the subscripts. So that means ##\epsilon_{123} = 1##. So how do you keep track of which is which? When I was younger, it was introduced to be a permutation, and a bunch of other nomenclature that I couldn't quite grasp until I was older. So I'll give you the secret I used! I simply start with 123, and then bump up the numbers to get 312, then again to get 231. Then, to get the -1, you just start with 321, and do the same. This is essentially what a permutation is, but when you're seeing all this for the first time, and a new symbol, even the simple things seem hard! So, how exactly does this relate to cross product? Well, let me throw out this equation (Note I am using Einstein summation): $$\vec{A} \times \vec{B} = \epsilon_{ijk}A^jB^k\hat{e_i}$$ Pretty obvious now, right? Okay, if it isn't obvious how they're related, don't worry! ##A^j## ##B^k## are the components of our vector, and ##\hat{e_i}##is our basis vector. Let's explicitly write out the sum to see if it will give us the same result as the cross product. \\
$$
\epsilon_{ijk}A^jB^k\hat{e_i} = $$
\begin{align*}
\epsilon_{111}A^1B^1\hat{e_1} + \epsilon_{112}A^1B^2\hat{e_1} + \epsilon_{113}A^1B^3\hat{e_1} + \\ \epsilon_{123}A^2B^3\hat{e_1} + \epsilon_{122}A^2B^2\hat{e_1} + \epsilon_{121}A^2B^1\hat{e_1} +\\
\epsilon_{131}A^3B^1\hat{e_1}+\epsilon_{132}A^3B^2\hat{e_1}+\epsilon_{133}A^3B^3\hat{e_1} +\\ \epsilon_{211}A^1B^1\hat{e_2}+ \epsilon_{212}A^1B^2\hat{e_2} + \epsilon_{213}A^1B^3\hat{e_2} + \\
\epsilon_{223}A^2B^3\hat{e_2} + \epsilon_{222}A^2B^2\hat{e_2} + \epsilon_{221}A^2B^1\hat{e_2} + \\
\epsilon_{231}A^3B^1\hat{e_2} + \epsilon_{232}A^3B^2\hat{e_2} + \epsilon_{233}A^3B^3\hat{e_2} +\\
\epsilon_{311}A^1B^1\hat{e_3} + \epsilon_{312}A^1B^2\hat{e_3} + \epsilon_{313}A^1B^3\hat{e_3} + \\
\epsilon_{323}A^2B^3\hat{e_3} + \epsilon_{322}A^2B^2\hat{e_3} + \epsilon_{321}A^2B^1\hat{e_3} +\\
\epsilon_{331}A^3B^1\hat{e_3} + \epsilon_{332}A^3B^2\hat{e_3} + \epsilon_{333}A^3B^3\hat{e_3}
\end{align*}
So, I know what you're thinking... 27 terms from such a simple line?! That's the power of the Einstein summation convention! However, I only wrote this out so you can get a feel for that power. In reality, we just use what the Levi-Civita symbol tells, that is any ##i=j=k## just go to zero, so our 27 equations zoom down to 6.
So we're left with
\begin{align*}
\epsilon_{ijk}A^jB^k\hat{e_i} =
\epsilon_{123}A^2B^3\hat{e_1} + \epsilon_{132}A^3B^2\hat{e_1} \\
\epsilon_{231}A^3B^1\hat{e_2} + \epsilon_{213}A^1B^3\hat{e_2} \\
\epsilon_{312}A^1B^2\hat{e_3} + \epsilon_{321}A^2B^1\hat{e_3}
\end{align*}
I've specifically put them in this order for a reason, we can now factor out the specific basis vector on each line and let's see what we get.
We now see that our equation becomes
\begin{align*}
\epsilon_{ijk}A^jB^k\hat{e_i} =
\hat{e_1}(A^2B^3 - A^3B^2) + \\
\hat{e_2}(A^3B^1 - A^1B^3)+ \\
\hat{e_3}(A^1B^2 - A^2B^1)
\end{align*}Which, as you can see, makes no mention of the particular coordinate system you're in. I do go into more details in my notes, but I think this is long enough for you to get the idea!