Two scenarios should be considered for this problem:

First scenario: x=2, y=3 and z=4 (so you have two even and one odd number)
Second scenario: x=1, y=2 and z=3 (the sequence contains two odd and one even number)

For (x+y)(y+z) both scenarios result in an odd integer.

yes, and C is not correct, because even*odd+even gives even

Here is my approach

I used 3 different values of x,y, and z.There is no point in continuing further because these are 3 consecutive integers and after this the pattern (of odds and evens will repeat).

x y z xyz x+y+z xy+z (x+y)z (x+y)(y+z)

1 2 3 6 6 5 (Eliminate A and B)
2 3 4 10 20 35 (Eliminate C and D)
3 4 5


Therefore, by the principal of elimination the answer is E