This is how I solved.
f(abc) = 3 * f(def)
So, 5^a*2^b*3^c = 3*[5^d*2^e*3^f]
So, (5^a*2^b*3^c)/(5^d*2^e*3^f) = 3
So, 5^(a-d)*2^(b-e)*3^(c-f) = 3^1
So,5^(a-d)*2^(b-e)*3^(c-f) = 3^1*5^0*2^0
S0, a-d =0, b-e =0 and c -f =1.
So, a =d, b = e and c = f +1
So abc - def is equal to abc - abf
And since c = f + 1, differrence is 1
Answer is A

Incognito,

Could you please detail your work?

Thanks

Sure. The function consists purely of powers of 5, 2 and 3. None of which are multiples of each other.

Since f(abc) = 3*f(def), and since the function is entirely made up of multiples of 5, 2 and 3, it would imply that the only difference is a power of 3. This would mean that a = d (since there is no multiple of 5 from the functions output), b = e (no multiple of 2 in the functions output) and c = f + 1 (since there is a multiple of 3 in the two functions output). Since abc and def are 3-digit numbers, and c = f + 1, the difference between abc and def would be 1. Hope that helps!
_________________

My GMAT debrief

I went totally tangent on this one.

thought abc is a*b*c and not a number abc

and wasnt sure why every one was thinking its too obvious.

lesson learnt!

oh yeah..tricky one
f(def) = 5^d*2*e*3^f
a is the same multiplied by 3
we have same prime factorization
abc = def+1
abc-def = 1

f(abc) = 5^a * 2^b *3^c
f(def) = 5^d * 2^e * 3^f

since, f(abc) = 3*f(def)
5^a * 2^b *3^c = 5^d * 2^e * 3^(f+1)

so, a=d
b=e
c=f+1

so, for example, if abcis 123, then def is 124,

hence difference is 1.