Exactly one of the following groups of inequalities is impossible — no values of the variables can satisfy every relation in the group at the same time. Which group is it?

Join the relations end to end and see whether the loop closes on a contradiction. In the correct option, A > B, B ≥ C and C > A compose to A > B ≥ C > A, that is A > A, which no number satisfies — the group is impossible. The others all close harmlessly. 'A > B, B ≥ C, A > C' is satisfied by A = 5, B = 3, C = 3, and the third relation is merely the consequence of the first two. 'A ≥ B, B ≥ C, C = A' is the tempting one, because it also loops back to A, but every link is non-strict, so the loop just forces A = B = C, which is perfectly consistent. 'A < B, B ≤ C, A < C' is satisfied by A = 1, B = 4, C = 6.