I also found the same answer as Karan
2 -2 2 | 2
0 -2 3 | 1
0 5 -2 | 3
----------------
2 1 3
My method was as follows
From condition 3, we can parameterise the matrix with 5 variables
. . A | r1
. . B | r2
E D C | r3
----------------
c1 c2 c3
Since C occurs r3 times in the grid (rule 1) and c3 times in the grid (rule 2), it follows that r3 = c3.
Since there are 9 squares to fill in the grid, and each number in the grid is at the end of a row or column, $r_1 + r_2 + r_3 + c_1 + c_2 = 9$. As each row and column ends with a number, the total for that row or column must be at least 1. $r_1, r_2, r_3, c_1, c_2 >= 1$.
Searching for possible values for $r_1$ through $c_2$ yields ~$70$ results.
(1, 1, 1, 1, 5), (1, 1, 1, 5, 1), (1, 1, 1, 2, 4), $\dots$
As the sum of the rows is the same of as the sum of the columns, $r_1 + r_2 + r_3 = c_1 + c_2 + c_3$, and $r_3 = c_3$, the first two rows and columns sum to the same value. $r_1 + r_2 = c_1 + c_2$. We can ignore any permutation of (1, 1, 1, 2, 4) as there is no way to assign the numbers so that $r_1 + r_2 = c_1 + c_2$. We can discard any permutation that does not satisfy that above constraint.
As we can make a trivial new solution by swapping the first and second row, the first and second column, and transposing the matrix, we can w.l.o.g. reduce the possible row and column totals down to 5 general cases
. . A | 1
. . B | 1
E D C | 5
----------------
1 1 5
. . A | 3
. . B | 1
E D C | 1
----------------
3 1 1
. . A | 3
. . B | 1
E D C | 1
----------------
2 2 1
. . A | 2
. . B | 2
E D C | 1
----------------
2 2 1
. . A | 2
. . B | 1
E D C | 3
----------------
2 1 3
Next I bruteforced
The first case has one possiblity which immediately leads to a contradiction. Each dot becomes a C, and comparing the first two rows $C + C + A = 1 = C + C + B$ leads to the conclusion that A and B are not distinct, contradicting rule 3.
Searching through each possible assignment of variables for viable matrices gave:
$r_1, r_2, r_3, c_1, c_2$: viable / possible
1, 1, 5, 1, 1: 0 / 1
3, 1, 1, 1, 3: 0 / 6
1, 1, 5, 1, 1: 0 / 12
2, 2, 1, 2, 2: 2 / 24
2, 1, 3, 1, 2: 4 / 24
However half of these are actually the same matrix transposed and with the variables relabelled. So the 3 possible assignments of variables are
E A A | 2
B D B | 2
E D C | 1
----------------
2 2 1
This one can actually be ruled out by summing the first two rows and columns, and noticing that it cannot produce integer values for the variables.
$E + A + A + B + D + B + E + B + E + A + D + D = 3(A + B + D + E) = 8$.
Potentially there are some answers with rational fraction solutions, but I did not look for any.
A C A | 2
E C B | 1
E D C | 3
----------------
2 1 3
C C A | 2
E A B | 1
E D C | 3
----------------
2 1 3
Since these are systems of linear equations, I could solve them to get the original solution, and the other solution given in the linked answer. So I'm pretty sure these are the only two integral solutions.