Question 847878
Let the consecutive integers be known as: n, n+1, and n+2.

We know that n(n+2) - (n+1) = 1 + 6*(n+2)

n^2 + 2n - n - 1 = 1 + 6n + 12

n^2 + n -1 = 6n + 13

n^2 -5n -14 = 0

(n-7)(n+2) = 0

n = 7

Then the numbers are 7, 8 ,9... 7 being the smallest.