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Let $B = \begin{bmatrix} x & y & z \end{bmatrix}$. Since $A$ is a $3 \times 1$ matrix and $B$ is a $1 \times 3$ matrix, the product $AB$ will be a $3 \times 3$ matrix.
We have $$AB = \begin{bmatrix}1\\ 4\\ -2\end{bmatrix} \begin{bmatrix} x & y & z \end{bmatrix} = \begin{bmatrix} x & y & z \\ 4x & 4y & 4z \\ -2x & -2y & -2z \end{bmatrix}$$
We are given that $AB = C$, where $C = \begin{bmatrix}3&4&2\\ 12&16&8\\ -6&-8&-4\end{bmatrix}$. Therefore, we have $$\begin{bmatrix} x & y & z \\ 4x & 4y & 4z \\ -2x & -2y & -2z \end{bmatrix} = \begin{bmatrix}3&4&2\\ 12&16&8\\ -6&-8&-4\end{bmatrix}$$
By comparing the entries of the matrices, we get the following equations: $x = 3$ $y = 4$ $z = 2$ $4x = 12$ $4y = 16$ $4z = 8$ $-2x = -6$ $-2y = -8$ $-2z = -4$ The first three equations give us $x = 3$, $y = 4$, and $z = 2$. The other equations are consistent with these values.
Therefore, $B = \begin{bmatrix} 3 & 4 & 2 \end{bmatrix}$.
Final Answer: $\begin{bmatrix} 3 & 4 & 2 \end{bmatrix}$