Question 663691
Let x be the first integer.  Then the next two consecutive integers are x+1 and x+2.  
Three times the sum of all three is 3*(x + (x+1) + (x+2))
The product of the larger two is (x+1) * (x+2)
So we have
(x+1) * (x+2) = 3*(x + (x+1) + (x+2))
x^2 + 3x + 2 = 3*(3x +3)
x^2 + 3x + 2 = 9x + 9
x^2 + 3x + 2 - 9x - 9 = 9x + 9 - 9x - 9
x^2 - 6x - 7 = 0
We can solve for x using the quadratic equation {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} with a=1, b=-6 and c=-7
{{{x = (-(-6) +- sqrt( (-6)^2-4*1*(-7) ))/(2*1) }}}
{{{x = (6 +- sqrt( 36+28 ))/2 }}}
{{{x = (6 +- sqrt( 64 ))/2 }}}
{{{x = (6 +- 8)/2 }}}
{{{x = 14/2 = 7}}} or {{{x = -2/2 = -1}}}
So there are two solutions to this problem:
7, 8, 9 and -1, 0, 1