Let's start by representing the three consecutive even integers as x, x+2, and x+4, where x is the smallest of the three.
According to the problem, the product of the first and second is 80 more than 15 times the third.
We can translate this into an equation as follows:
x(x+2) = 15(x+4) + 80
Expanding the left side and simplifying,
we get:
x^2 + 2x = 15x + 100
Bringing all the terms to one side,
we get:
x^2 - 13x - 100 = 0
We can solve this quadratic equation using the quadratic formula:
x = [13 ± √(13² + 4(100))]/2
x = [13 ± √729]/2
x = [13 ± 27]/2
We reject the negative solution and take the three consecutive even integers as:
x = 20, x+2 = 22, x+4 = 24
Therefore, the three positive consecutive even integers that satisfy the given condition are 20, 22, and 24.