Answer (C): Let $a_{ij}$ be the integer at row $i$ and column $j$. It is given that $a_{55} = 0$, $a_{24} = 48$, $a_{43} = 16$, and $a_{31} = 12$. Suppose $a_{54} = d$. Then row $5$ is $4d, 3d, 2d, d, 0$ because it is an arithmetic progression with common difference $-d$. The arithmetic progression in column $1$ gives
$$a_{41} = \frac{a_{31} + a_{51}}{2} = \frac{12 + 4d}{2} = 6 + 2d.$$

Row $4$ gives
$$a_{43} = 16 = \frac{2a_{44} + a_{41}}{3} = \frac{\frac{4}{3}d + 32 + 6 + 2d}{3},$$
which implies $48 = \frac{10}{3}d + 38$, so $d = 3$. Filling in column $3$ with common difference $16 - 6 = 10$ and column $1$ with difference $12 - 12 = 0$ produces $a_{13} = 46$ and $a_{11} = 12$. Finally,
$$a_{12} = \frac{a_{13} + a_{11}}{2} = \frac{46 + 12}{2} = 29.$$

The full array looks like this:
$$\begin{bmatrix} 12 & \underline{\mathbf{29}} & 46 & 63 & 80 \\ 12 & 24 & 36 & \mathbf{48} & 60 \\ \mathbf{12} & 19 & 26 & 33 & 40 \\ 12 & 14 & \mathbf{16} & 18 & 20 \\ 12 & 9 & 6 & 3 & \mathbf{0} \end{bmatrix}$$