If a, b and c are the increasing terms of a G.P. with common ratio r, given…

For three terms a, b, c in a G.P. with common ratio r: b = a × r and c = a × r2, so a × c = b2 -- the product of the outer terms always equals the square of the middle term, and therefore a × b × c = b3.

Applying this to the given condition:

1. (log a + log b + log c) / log 6 = 6, so log(abc) / log 6 = 6, so log6(abc) = 6, so abc = 66.

2. Since abc = b3 for a G.P., b3 = 66 = (62)3, so b = 62 = 36.

3. The G.P. is increasing, so c = b × r with r > 1, giving c − b = 36r − 36 = 36 × (r − 1), which is positive but has no built-in lower bound unless the terms are restricted further.

4. For a, b, c to all be whole numbers (standard for this kind of numeric G.P. puzzle), a = b2 / c = 1296 / c must also be a whole number, so c must be a divisor of 1296 = 24 × 34 that is larger than b = 36.

5. The divisors of 1296 just above 36 are 48, 54, 72, and so on; the smallest is 48.

6. Minimum (c − b) = 48 − 36 = 12.

Check: with c = 48, a = 1296 ÷ 48 = 27, giving the whole-number G.P. 27, 36, 48 (ratio 4/3), and 27 × 36 × 48 = 46656 = 66, confirming the condition holds.