1. Analyze the Equation's Structure:

   The equation is ABA * AA = ACCA. The last digit of the product is determined by the last digits of the numbers being multiplied. Here, the last digit of (A * A) must be A.

   We can test the digits from 1 to 9 to see which ones satisfy this condition:

   • 1 * 1 = 1 (A=1 is possible)
   • 5 * 5 = 25 (A=5 is possible)
   • 6 * 6 = 36 (A=6 is possible)

   So, A can only be 1, 5, or 6.

2. Test the Possibilities for A:
   • If A = 5, the equation is 5B5 * 55 = 5CC5. The result of 500+ * 55 will be a 5-digit number, but 5CC5 is a 4-digit number. This is impossible.
   • If A = 6, the equation is 6B6 * 66 = 6CC6. The result will also be a 5-digit number. This is also impossible.

   Therefore, the only possibility is A = 1.

3. Solve for the case where A = 1:

   The equation becomes 1B1 * 11 = 1CC1. We can express this algebraically:

   (101 + 10*B) * 11 = 1001 + 110*C 1111 + 110*B = 1001 + 110*C 110 + 110*B = 110*C

   Dividing the entire equation by 110 gives us the simple relationship: 1 + B = C.

4. Find All Valid Solutions and their Sums:

   We need to find pairs of B and C where C = B + 1, and A, B, and C are all different digits from 1 to 9. Since A=1, B cannot be 1.

   • If B=2, C=3 → (1,2,3) → Sum = 6
   • If B=3, C=4 → (1,3,4) → Sum = 8
   • If B=4, C=5 → (1,4,5) → Sum = 10
   • If B=5, C=6 → (1,5,6) → Sum = 12
   • If B=6, C=7 → (1,6,7) → Sum = 14
   • If B=7, C=8 → (1,7,8) → Sum = 16
   • If B=8, C=9 → (1,8,9) → Sum = 18

Conclusion: The possible values for the sum (A+B+C) are {6, 8, 10, 12, 14, 16, 18}. These are 7 different, unique values.

Final Answer: The correct option is C. 7.