If we place 192 in the 1st row, 384 in the second row and 576 in the 3rd row, then we get a 3x3 square, exhausting all non-zero digits.
In this square 1st row + 2nd row = 3rd row.
Please provide all other arrangements, satisfying this condition.
By the way:
The 3rd row must be a multiple of 9.
Let a = first row, b = 2nd row, c = 3rd row.
By casting out 9's,
a + b + c = 0 (mod 9) (because all digits are there exactly once)
Also, a + b - c = 0 (initial problem)
Subtracting the two equations, we get
2c = 0 mod 9,
So c = 0 mod 9