Let the given matrix be C. We are given that A is a symmetric matrix (A^T = A) and B is a skew-symmetric matrix (B^T = -B). The sum of these matrices is A+B = C.

Step 1: Let the given matrix be $C$. We are given that $A$ is a symmetric matrix ($A^T = A$) and $B$ is a skew-symmetric matrix ($B^T = -B$). The sum of these matrices is $A+B = C$.
The given matrix $C$ is:

$$C = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$$

The transpose of $C$ is $C^T$:

$$C^T = \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}$$

Step 2: We know that $A = \frac{1}{2}(C + C^T)$.
First, calculate $C + C^T$:

$$C + C^T = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} + \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix} = \begin{bmatrix} 2+2 & -2-1 & -4+1 \\ -1-2 & 3+3 & 4-2 \\ 1-4 & -2+4 & -3-3 \end{bmatrix} = \begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix}$$

Now, calculate $A$:

$$A = \frac{1}{2} \begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix} = \begin{bmatrix} 2 & -3/2 & -3/2 \\ -3/2 & 3 & 1 \\ -3/2 & 1 & -3 \end{bmatrix}$$

Step 3: We know that $B = \frac{1}{2}(C - C^T)$.
First, calculate $C - C^T$:

$$C - C^T = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} - \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix} = \begin{bmatrix} 2-2 & -2-(-1) & -4-1 \\ -1-(-2) & 3-3 & 4-(-2) \\ 1-(-4) & -2-4 & -3-(-3) \end{bmatrix} = \begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix}$$

Now, calculate $B$:

$$B = \frac{1}{2} \begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1/2 & -5/2 \\ 1/2 & 0 & 3 \\ 5/2 & -3 & 0 \end{bmatrix}$$

Step 4: Evaluate the product $AB$.

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