One idea for finding the solution could involve prime factorizations of pandigitals.  We can get an idea of the largest prime number which could be a factor of A, B, and C.  We are looking for the smallest N.  And the largest of A,B,C must be (see my earlier note) less than 8187654240 (the closest pandigital that is smaller is 8179654320).

If we imagine the prime factorizations of A, B, and C are:

3*3*x*y*p, 3*3*x*x*p, and 3*3*y*y*p,

(since all pandigitals are multiples of 9)

and the smallest primes x and y can be are 2 and 3,

then p_max satisfies:  8179654320 = 3*3*3*3*p_max

p_max = 8179654320 / 81 = 100983386.6666  

(actually the next smaller prime, which is 100983367)

So if looking at prime factorizations of A,B,C to find triplets which when combined have prime factors in multiples of 3, no primes over 100983367 need be considered.