Question 586422
Let the integers be n, n+1, and n+2.
The product of the first two is {{{n(n+1)=n^2+n}}}
That plus the third integer is {{{n^2+n+n+2=n^2+2n+2}}}
If {{{n^2+2n+2=101}}} --> {{{n^2+2n=99}}} (subtracting 2 from both sides).
Now we can complete the square to solve the quadratic equation:
{{{n^2+2n=99}}} --> {{{n^2+2n+1=99+1}}} --> {{{(n+1)^2=100}}} --> {{{n+1=10}}}
The integers are 9, 10, and 11.
{{{9*10+11=90+11=101}}}
NOTE: There are other ways to solve that quadratic equation (factoring, the quadratic formula), but completing the square looked easier.