In the magic square, the sum of the three numbers in any row, column or diagonal is the same. The sum of the three numbers in any row is
(A) 0
(B) 1
(C) 3
(D) 7
(E) 9

| $2 x$ | 3 | 2 |
| :---: | :---: | :---: |
|  |  | -3 |
| 0 | $x$ |  |
The sum of the numbers in the $1^{\text {st }}$ row is $2 x+5$, so the sum of the numbers in any row, column or diagonal is also $2 x+5$.

Therefore, the entry in the $2^{\text {nd }}$ row, $1^{\text {st }}$ column must be 5 , using the $1^{\text {st }}$ column.

Thus, the entry in the $2^{\text {nd }}$ row, $2^{\text {nd }}$ column is $2 x+3$, using the $2^{\text {nd }}$ row.

Looking at the sum of the $2^{\text {nd }}$ column,

$$
\begin{aligned}
3+(2 x+3)+x & =2 x+5 \\
3 x+6 & =2 x+5 \\
x & =-1
\end{aligned}
$$

so the sum of the numbers in any row is $2(-1)+5=3$.

Answer: (C)