100 each

First of all, Note that the number of plates left for group C is an even number and hence can be representated as $2k$ for some 'k'.
Now, $\frac{1}{4}c = 2k$ ;
==> $c = 8k$ .
So, number of people of group C is a multiple of 8.
Now, note that for $c\geq 80$ , we have average number of plates left $\geq 4$ that is either all remaining members are of group B or else, its impossible. We dont want either case since, we must have at least 1 member of each group.

Now, for $c\leq 56$ , we have average number of plates left $<2$ and hence the required number is
$>56 ; <80$ and a multiple of 8.
There remain just 2 options:

CASE 1:
$c = 64$
We have;
$a+b = 36$
$2a + 4b = 84$
Solving we get; $a=30; b=6$

CASE 2:
$c = 72$
We have;
$a+b = 28$
$2a + 4b = 82$
Solving, we get; $a=15 ; b=13$

Hence the answer is :
$(72*15*13)-(64*30*6) = 14040 - 11520 = \boxed{2520}$

We have $a+b+c=100$ (1)

And $2a+4b+\frac{c}{4}=100$ (2)

Multiplying (2) by 4 we get $8a+16b+c=400$ (3)

Subtracting (1) from (3) we get $7a+15b=300$ (4)

Now as 300 and $15b$ are both multiples of 15 so must be $7a$ and thus $a$ must be a multiple of 15

However if $a\geq45$ then $b$ ends up negative so we have either $a=15$ or $a=30$ which from (4) give $b=13$ or $b=6$ respectively which from (1) give $c=72$ or $c=64$ respectively.

So $abc=15\times13\times72=14040$ or $abc=30\times6\times64=11520$ whose difference is $\fbox{2520}$