Question 252073
Let n be the first integer, n+1 be the second, and n+2 be the third integer. The sum of these three is n+n+1+n+2 = 3n + 3. This sum must equal 6. So, we have

n+n+1+n+2 = 6
3n + 3 = 6
3n = 3
n = 1

We have our three consecutive integers in order as 1, 2, 3.