Triple the fun ...

A quick overview of one possible method ....

Let $x = a + 2, y = b + 2$ and $z = c + 2$ . Then the given equation becomes

$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{2}$ .

Now look for solutions with $3 \le x \le 6$ and $x \le y \le z$ . A straightforward search gives us the following triples $(x,y,z)$ :

$(3,7,42), (3,8,24), (3,9,18), (3,10,15), (3,12,12), (4,5,20), (4,6,12), (4,8,8), (5,5,10)$ and $(6,6,6)$ .

This translates to the following triples $(a,b,c)$ :

$(1,5,40), (1,6,22), (1,7,16), (1,8,13), (1,10,10), (2,3,18), (2,4,10), (2,6,6), (3,3,8)$ and $(4,4,4)$ .

Factoring in all distinct permutations of these $10$ triples gives us a final number of $\boxed{46}$ ordered triples satisfying the original equation.