Let the number formed in the first row be X. The numbers formed in the second and third rows are 2X and 3X respectively. All digits from 1 to 9 must be used exactly once across the three rows. The sum of digits 1 through 9 is 45. The sum of the numbers X, 2X, and 3X is 6X. For X to be a three-digit number, 2X and 3X must also be three-digit numbers. The sum of the digits of X, 2X, and 3X, when treated as numbers, must be divisible by 9 if the digits are a permutation of 1-9. However, the condition is that the digits themselves are unique and form these numbers. The sum of the digits used (1-9) is 45. The sum of the three numbers (X + 2X + 3X) must be 6X. If X is a three-digit number, the smallest possible sum of digits from 1-9 is 1+2+3 = 6 and the largest is 7+8+9 = 24 for each row. The total number formed by concatenating the digits of X, 2X, and 3X must be a permutation of 1-9. Considering the constraints and the possible values for X such that 2X and 3X are also 3-digit numbers using unique digits from 1-9, only specific sets of X are possible. The problem further restricts the digits available for the first row (X) to a subset of {2, 3, 7, 9}. By testing these combinations, it's found that only two valid arrangements satisfy all conditions.