abcdecgf

There are 6 unique letters, one is repeated at positions 3 and 6.

There must be 7! (No, not seven factorial...)

I don't know what the word is, so the letter "c" is only an example.

Are the positions fixed too? I'm assuming they are, since the problem is slightly more complex like that. Of course there are tons of ways to do it. And as shmem wrote, probably not best done with a single regex - although it may be possible, perhaps by means of one of those funky extensions still marked as "experimental". One possible way that springs to my mind is:

#!/usr/bin/perl

use strict;
use warnings;

$_=<<'.';
bEjhMELGUaL smtMDEYSxyDvuQiUfAbJfYMPnfJAqaPnKL VWZWSdfYRSaSGlXOyPfxusC
dtRAHabcdecgf taNdvtKdBlJcnFryVXObEDvawRyviWO hwlKiBpDWYeBPYhlpKFvrSeQ
ksWmkXqQdLQPIzvKFE Jqrclq mPqQbMvkAx LtVuFMehKirSATuqlFzqwRknocsrcKXAE
FNbOivdvkRonEkg apuPyHpTlssvVs BbwiHBvhfrSFwVkhwHkvoYjaGgntzFbEvPCIttD
IAlYqoLUjtxsYvbwUBHIoMYmPJbGeXymuwERkHwSyKbE XMCcFgsYPzmJbVUsOwfDTgUiJ
.

while (/(?=([a-zA-Z]{8}))/g) {
    my @l = (my $found=$1) =~ /./g;
    print $found, "\n" if $l[2] eq $l[5] and do {
        my %h; @h{@l}=(); 7 == keys %h;
    }
}
  
__END__
[download]

Update: or, in a slightly more agile way:

for (/(?=([A-zA-Z]{8}))/g) {
    my @l=/./g;
    print $_, "\n" if $l[2] eq $l[5] and do {
        my %h; @h{@l}=(); 7 == keys %h;
    }
}
[download]