1/a + 1/b + 1/c = 3/4 how many ordered triplets of (a,b,c) are there??

Just a ask here, do we take a,b,c as to always >0, there is a possibility of set having (1,-2,4) appears as well.

Let us try to enumerate all the cases and count the number of triplets. 

Without loss of generality, we can assume that a <= b <= c.

Therefore, the maximum value 'a' can take is 4.

In the case where a = 2, 1/b + 1/c = 1/4 or c = 4*b/(4-b)
The only possible natural number solutions to this are b = 5, c = 20; b = 6, c= 12 and b = c = 8

In the case where a = 3, 1/b + 1/c = 5/12 or c = 12*b/(5b-12)
The only possible natural number solutions to this are b = 3, c = 12 and b = 4, c= 6

In the case where a = 4, 1/b + 1/c = 1/2 and the only possible case is a=b=c=4

So, let us write all the possible triplets again (2,5,20); (2,6,12); (2,8,8); (3,3,12); (3,4,6) and (4,4,4)

Hence, the number of triplets possible is 6 + 6 + 3 + 3 + 6 + 1 = 25