Comparing all elements with all elements would be quadratic. Here's a solution that runs in time O (n log n + k) where n is the number of rows and k the number of pairs for which the fields 0, 2 and 3 are equal.

This could still be inefficient, k could be n^2/2 but still nothing is reported because we only need to report the pairs for which field 1 differs. OTOH, if it's known that if the fields 0, 2, and 3 are equal that then field 1 is different, the algorithm used is optimal:

use strict;
use warnings;

my @AoA;

while (<DATA>) {
    push @AoA => [split];
}

my @sort = sort {$a -> [0] cmp $b -> [0]  ||
                 $a -> [2] <=> $b -> [2]  ||
                 $a -> [3] <=> $b -> [3]} @AoA;

foreach my $i (0 .. $#sort - 1) {
    foreach my $j ($i + 1 .. $#sort) {
        last unless $sort [$i] -> [0] eq $sort [$j] -> [0] &&
                    $sort [$i] -> [2] == $sort [$j] -> [2] &&
                    $sort [$i] -> [3] == $sort [$j] -> [3];
        next if $sort [$i] -> [1] eq $sort [$j] -> [1];
        print "[@{$sort[$i]}] and [@{$sort[$j]}]\n";
    }
}

__DATA__
AAA BUY 98 0
BBB SEL 27 1
FFF BUY 43 4
AAA SEL 98 0
CCC SEL 98 0
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Abigail