Yes, she is right.

Suppose, on the contrary, that A, B and C made no mistakes. They all obtained different answers, so they must all have chosen different pairs of numbers from the set {x, y, z}. But if you take the HCF of two different pairs from a triple of numbers, and then take the HCF of those two HCFs, you will get the HCF of the original triple of numbers.

If that's not clear, look at an example where the numbers are small. Suppose you wanted to find the HCF of 12, 20 and 30. You could start by finding that HCF(12,20) = 4, HCF(12,30) = 6 and HCF(20,30) = 10. If you take any pair from the numbers 4, 6 and 10, you will find that their HCF is 2. (So HCF(12,20,30) = 2.)

Coming back to the problem with x, y and z, we are told that the three HCFs of pairs of these numbers are 1000004, 1000006 and 1000008. However, HCF(1000004, 1000006) = HCF(1000006, 1000008) = 2 but HCF(1000004, 1000008) = 4. Since 4 is different from 2, someone somewhere must have made a mistake.