The solutions already given are very compact and it seems to me that this is cluttering and obscuring a nice solution. As a matter of fact, I don't really understand the solutions given. (So that's my problem you say.) So I come up with an other solution, which to my is much clearer. This is it:

#! /usr/local/bin/perl

@coord = ( 6, 2, -4 );
$dim   = scalar ( @coord );

@trans = map { $_ - 1 } @coord;

for ( $i = 0; $i < 3**$dim; $i++ ) {
    @vector = vector ( $i );
    @add    = add_vector ( \@trans, \@vector );
    print "[@add]\n" unless center ( @vector );
}

sub vector {
    my $i = shift @_;
    my @vector;

    for ( my $j=0; $j < $dim; $j++ ) {
        push ( @vector, $i % 3 );
        $i = int $i/3;        
    }
    
    return @vector;
}

sub add_vector {
    my $ar_ref1 = shift @_;
    my $ar_ref2 = shift @_;
    my @vector;
    
    for ( my $j=0; $j < $dim; $j++ ) {
        push ( @vector, $$ar_ref1[$j] + $$ar_ref2[$j] );
    }
    
    return @vector;
}

sub center {
    my @vector = @_;
    my $bool   = 1;
    my $i      = 0;
    
    while ( $bool and $i < $dim ) {
        $bool = ( $vector[$i++] == 1 );
    }

    return $bool;
}
[download]

with the given coordinates it produces the following output

[5 1 -5]
[6 1 -5]
[7 1 -5]
[5 2 -5]
[6 2 -5]
[7 2 -5]
[5 3 -5]
[6 3 -5]
[7 3 -5]
[5 1 -4]
[6 1 -4]
[7 1 -4]
[5 2 -4]
[7 2 -4]
[5 3 -4]
[6 3 -4]
[7 3 -4]
[5 1 -3]
[6 1 -3]
[7 1 -3]
[5 2 -3]
[6 2 -3]
[7 2 -3]
[5 3 -3]
[6 3 -3]
[7 3 -3]
[download]

I am not going to explain my code and the idea's I used to create it. (It would be rather lengthy). The variables and subroutines I use suggest the underlying mechanisme. But if my code is as obscure to you as the others solutions are to my, please comment on the reply and I will try to explain it!