Sets of Glory

First we can find all the sets.( Assuming $a \le b \le c$ )
We can consider how small could $a$ be ?
First, set $a = 1$ . This will not work because it nothing for $\dfrac{1}{b} and \dfrac{1}{c}$ .
Set $a = 2$ and try values for $b$ ( with $b$ is greater than 2) :
If $b = 2$ , then $\dfrac{1}{c} = 0$ , which isn't good.
If $b = 3$ , then $c = 6$ , which works.
If $b = 4$ , then $c = 4$ , which works.
Then, set $a = 3$ and try the values for $b$ :
If $b = 3$ , then $c = 3$ , which works.
Finally, If $a = 4$ , then $b$ and $c$ should be $\le a$ , so we no need to look further.
So, we found that the sets are $(2, 3, 6)$ , $( 2, 4, 4)$ and $(3, 3, 3)$ .
So, the sum of a, b and c of the sets are :
2 + 3 + 6 = 11
2 + 4 + 4 = 10
3 + 3 + 3 = 9

Then, the largest is 11.