Weight distribution of the ternary extended Golay code $\mathcal G_{12}$

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Let $\mathcal G_{12}$ denote the extended ternary $[12, 6, 6]_3$ Golay code. I'm now asked to find the weight distribution of $G_{12}$. As a hint, I'm told I could utilize the MacWilliams identity (in the form that can be for example found here) at some point.

My attempt so far: I already know the most commonly known properties of $\mathcal G_{12}$, like that it's perfect, linear and self-dual, and that it's weight is $6$. Let $A_i$ denote the amount of code words of $G_{12}$ of weight $i$. I've already figured out that we have $A_i = 0$ for $3 \nmid i$ (as per self-duality), and we have $A_3 = 0$ because it is of weight $6$, and of course $A_0 = 1$ as $0 \in G_{12}$ because it's dual.

But from there on, I don't really know how to continue. I tried to brute-force the MacWilliams-identity onto it: per self-duality, we have, according to MacWilliams:

$$W_{\mathcal C}(X, Y) = W_{\mathcal C^\bot}(X, Y) = \frac 1{|\mathcal C|} W_{\mathcal C} (X + 2 Y, X - Y)$$

$$=> X^{12} + A_6 X^{6} Y^6 + A_9 X^3 Y^9 + A_{12} Y^{12} = \frac 1{3^6}( (X + 2Y)^{12} + A_6 (X + 2Y)^6(X - Y)^6 + A_9 (X + 2Y)^3 (X - Y)^9 + A_{12}(X - Y)^{12}) $$

But I'm not really sure what to do now. I mean, I could try to multiply out everything, but it looks like that's going to be so tedious and nasty that it just can't be the intended way to do it. Are there any clever ways to continue and to derive some facts about the $A_i$ without working my way through that entire thing?

I'm also not sure if that's already sufficient to figure out the weight distribution, as I haven't used much properties of the Golay code, except it's linearity and self-duality.

I'm of course also open to any other suggestions or solutions that don't use MacWilliams.

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It is a bit inconvenient to do by hand, but all you need are the terms $$ \begin{aligned} \frac1{729}W_{\mathcal{C}}(X+2Y,X-Y)&= \frac1{729}(1+a_6+a_9+a_{12})X^{12}+\\ &+\frac1{729}(24+6a_6-3a_9-12a_{12})X^{11}Y+\\ &+\frac1{729}(264 + 3 a_6 - 6 a_9+ 66 a_{12})X^{10} Y^2\\ &+\text{lower degree terms.} \end{aligned} $$

All you need to do is to compare those three coefficients to the three known corresponding coefficients of $W_{\mathcal{C}}(X,Y)$. Three equations, three unknowns. Routine.