Since it must be true for ANY three consecutive integers, 
it must be true for the SPECIFIC consecutive integers 1,2,3 
and also true for the SPECIFIC consecutive integers 2,3,4.  
So we will consider these two cases for each possible choice:

Case 1:  a=1, b=2, c=3   [Two odd numbers and one even number]
Case 2:  a=2, b=3, c=4   [Two even numbers and one odd number]

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(A) abc  

Case 1: abc = (1)(2)(3) = 6 which is an even number.

So (A) is ruled out just by Case 1. We don't even need to try Case 2.

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(B) a+b+c

Case 1: a+b+c = 1+2+3 = 6 which is an even number

So (B) is ruled out just by Case 1. We don't even need to try Case 2.

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(C) a+bc

Case 1: a+bc = 1+2×3 = 1+6 = 7, which is an odd number.
Case 2: a+bc = 2+3×4 = 2+12 = 14, which is an even number.

So (C) is ruled out by Case 2. 

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(D) a(b+c)

Case 1: a(b+c) = 1(2+3) = 1(5) = 5, which is an odd number.
Case 2: a(b+c) = 2(3+4) = 2(7) = 14, which is an even number.

So (D) is ruled out by Case 2. 

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(E) (a+b)(b+c)

Case 1: (a+b)(b+c) = (1+2)(2+3) = (3)(5) = 15, which is an odd number.
Case 2: (a+b)(b+c) = (2+3)(3+4) = (5)(7) = 35, which is an odd number.

So the only possible correct choice is (E).

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Edwin