Math algebra problem in junior high school (urgent ~)

Answer: Let the largest three digits be 100a+ 10b+c, and the smallest three digits be 100c+ 10b+a, and the two digits are subtracted = 99 (a-c) from1≤ c. Then 99 (a-c) = 198, 297, 396, 495, 594, 693, 792, these seven results, then there must be 9 in A, B and C, and the sum of the other two numbers =9, then the three numbers A, B and C can only be the above seven situations.

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I am a senior one student. In the experimental class, mathematics and physics are weak points. I often can't do the problems that others can do. I'm under a lot of pressure and I'm about to collapse.