Answer: A, B, C
In the isothermal path T is constant, so ΔU=0 and the gas absorbs heat equal to the work it does. In the adiabatic path q=0, so the gas does work at the expense of U and cools, giving ΔU<0 and less work than in the isothermal case. Hence ΔU differs between the paths, and statements 1, 2 and 3 are correct.