Don't Guess!

Factorization makes everything simple. The question, when factorized, reduces to:- $(a-1)(b-1)(c-1)=2012$ .

The factorization technique may be called Completing the Cuboid, for I could not find any other name... A close relative of SFFT, though!

So, the trick is:- $(a-1)(b-1)(c-1)=abc-ab-bc-ca+a+b+c-1$

And we subtract $1$ from both sides to complete the cuboid.

Since $a$ , $b$ , $c$ are positive integers, so are $a-1$ , $b-1$ and $c-1$ (they can''t be zero for their product is $2012$ ), because the sum and difference of $n$ integers is an integer.

So, we have $(a-1)(b-1)(c-1)=2012$

Number of positive integral factors of $2012=2^2 \times 504$ is $(2+1)(1+1)=6$ , which are $1$ , $2$ , $4$ , $503$ , $1006$ , and $2012$ .

The number of ways in which $2012$ can be written as a product of $3$ numbers is thus $18$ , which consequently gives $18$ ordered pairs of positive integers $(a,b,c)$ .

abc-ab-bc-ca+a+b+c-1 = 2012 = (a-1)(b-1)(c-1)
2012 = 2x2x503
(a-1)(b-1)(c-1) = (1x1x2012) , (1x2x1006), (2x2x503), (1x4x503)
=> there are 18 ordered triples