smc

If you change $3$ numbers, then you either change one number in each column and row (ie sudoku-style): $$\left[\begin{array}{ccc}\ast & 9& 2\\ 8& \ast & 6\\ 3& 5& \ast \end{array}\right]$$ Or you leave at least one row and one column unchanged: $$\left[\begin{array}{ccc}\ast & 9& 2\\ \ast & \ast & 6\\ 3& 5& 7\end{array}\right]$$ In the first case, you are changing just one common number in two sums, so you wind up with three pairs of sums. (In the example given, the sum in row $x$ is the same as in column $x$.) In the second case, since two of the sums are unchanged, and the sums started out equal, they must remain equal. (In the second example given, row $3$ and column $3$ are untouched.) Either way, $3$ changes is not enough. However, building on the second example, if you change either the untouched column or the untouched row, you will get a possible answer: $$\left[\begin{array}{ccc}\ast & 9& 2\\ \ast & \ast & 6\\ 3& \ast & 7\end{array}\right]$$ Letting the $\ast$ be a zero does indeed give $6$ different sums, so the answer is $4$, which is option $\boxed{4}$.