[THOUGHT] Let me analyze this problem step by step.

We have a family $\mathcal{F}$ of linear subspaces $C \subseteq \{0,1\}^{24}$ with:

1. Dimension 12
2. All weights divisible by 4
3. Contains the all-ones vector $\mathbf{1}$

The weight distribution of any $C \in \mathcal{F}$ can be expressed as a linear combination of weight distributions of two basis codes $C_1$ and $C_2$.

First, let me find the weight distribution of $C_1$ from its weight enumerator polynomial: $W_{C_1}(x, y) = (x^8 + 14x^4y^4 + y^8)^3$

Then I need to find the weight distribution of $C_2$, which has $A_4(C_2) = 0$ and $A_8(C_2) = 759$.

Since any code in $\mathcal{F}$ has its weight distribution as a linear combination of $C_1$ and $C_2$'s distributions, and $C^*$ has $A_4(C^*) = 6$, I can find the coefficients and then compute $A_{12}(C^*)$.

Let me start by computing the weight distribution of $C_1$. [/THOUGHT] [PYTHON] from sympy import symbols, expand, Poly

x, y = symbols('x y')

Weight enumerator for C1

W_C1 = (x8 + 14*x4*y4 + y8)**3 W_C1_expanded = expand(W_C1) print("W_C1 expanded:", W_C1_expanded)

Extract coefficients for each weight k (coefficient of x^(24-k) * y^k)

A_C1 = {} for k in range(0, 25, 4): # weights are multiples of 4 coeff = W_C1_expanded.coeff(x, 24-k).coeff(y, k) A_C1[k] = coeff print(f"A_{k}(C1) = {coeff}")

print("\nWeight distribution of C1:", A_C1) [/PYTHON]

[THOUGHT] Now I have the weight distribution of $C_1$. For $C_2$, I know that $A_4(C_2) = 0$ and $A_8(C_2) = 759$.

Since any code $C \in \mathcal{F}$ has weight distribution that is a linear combination of $C_1$ and $C_2$: $A_k(C) = \alpha \cdot A_k(C_1) + \beta \cdot A_k(C_2)$

For $C^*$ with $A_4(C^*) = 6$: $A_4(C^*) = \alpha \cdot 42 + \beta \cdot 0 = 6$ $\alpha = 6/42 = 1/7$

Since the total number of codewords is $2^{12} = 4096$ for any code in $\mathcal{F}$: $\alpha \cdot 4096 + \beta \cdot 4096 = 4096$ $\alpha + \beta = 1$ $\beta = 1 - 1/7 = 6/7$

Now I need to find $A_{12}(C_2)$. Using the constraint that $C_2$ also has 4096 codewords: $A_0(C_2) + A_4(C_2) + A_8(C_2) + A_{12}(C_2) + A_{16}(C_2) + A_{20}(C_2) + A_{24}(C_2) = 4096$

For $C_2$: $A_0(C_2) = 1$, $A_4(C_2) = 0$, $A_8(C_2) = 759$, $A_{24}(C_2) = 1$ (since $\mathbf{1} \in C_2$).

By symmetry (MacWilliams identity for self-dual codes), $A_k = A_{24-k}$:

• $A_{16}(C_2) = A_8(C_2) = 759$
• $A_{20}(C_2) = A_4(C_2) = 0$

So: $1 + 0 + 759 + A_{12}(C_2) + 759 + 0 + 1 = 4096$ $A_{12}(C_2) = 4096 - 1520 = 2576$

Now compute $A_{12}(C^*)$: $A_{12}(C^*) = \frac{1}{7} \cdot 2828 + \frac{6}{7} \cdot 2576 = \frac{2828 + 6 \cdot 2576}{7} = \frac{2828 + 15456}{7} = \frac{18284}{7} = 2612$ [/THOUGHT]

\boxed{2612}