First let us assign the three positive integers to be x, y, and z.
From the given problem statement, we know that:
(1/x) + (1/y) + (1/z) = 1
Without loss of generality we can assume x < y < z.
We know that:
1 = (1/3) + (1/3) + (1/3)
Where x = y = z = 3 would be a solution
However this could not be true because x, y, and z must all be different integers. And x, y, and z cannot all be 3 or bigger than 3 because the sum would then be less than 1. So let us say that x is a denominator that is less than 3. So x = 2, and we have:
(1/2) + (1/y) + (1/z) = 1
Therefore
(1/y) + (1/z) = 1/2
We also know that:
(1/4) + (1/4) = (1/2)
and y = z = 4 would be a solution, however this is also not true because y and z must also be different. And y and z cannot be larger than 4, so y=3, therefore
(1/2) + (1/3) + (1/z) = 1
Now we are left by 1 variable so we calculate for z. Multiply both sides by 6z:
3z + 2z + 6 = 6z
z = 6
Therefore:
(1/2) + (1/3) + (1/6) = 1
so {x,y,z}={2,3,6}