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nice question.... answer is A.
from stem we can infer that at least one from x,y,z is odd and at least one is even.
lets check st1. it is given that x(y+z) is even. can x be odd?
if it is, then (y+z) must be even, so they must have the same "oddity", and since there must be at least one even number, we can infer that y and z must be even if x is odd. however - this violates stem since in this arrangement xy+z is even.
so x cannot be odd - so it must be even. hence sufficient.
lets check st2
take x=1 y=2 z=3 and compare to y=1 x=2 z=3
both satisfy stem and st2, but in one x is odd and in the second x is even.