If i choose -1,0,1 as x,y,z respectively,only the 3rd choice would make sense , in which case the answer has to be option 3 only. please correct me if i am wrong.

Option I is xyz, so if you do (-1)(0)(1)=0 which is even, so I is true.

Option II says that x+y+z is even, son -1+0+1=0, which is even so II is true.

Option III (x+y)(y+z) so (-1+0)(0+1)=-1 is odd indeed, so III is true.

The correct anwer is E. I believe your problem is with the concept that 0 is an even integer.

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