It was pointed out that I may have misunderstood your concerns with my first reply. That you were probably alluding to the fact that the code I offered the OP only detects the first possible matching sequence, rather than returning all possible sequences. I attempted to offer the OP a starting point from which to understand how to develop recursive solutions, rather than just hand him a complete solution on a plate.

That starting point can easily be extended to return the numbers involved in the (first) sequence detected:

sub sumTo {
    my $target = shift;
    for my $i ( 0 .. $#_ ) {
        return [ $_[ $i ] ] if $_[ $i ] == $target;
        if( @{
            $_ = sumTo( $target - $_[ $i ], @_[ 0 .. $i-1, $i+1 .. $#_
+ ] )
        } ) {
            return [ $_[ $i ], @{ $_ } ];
        }
    }
    return [];
}
[download]

And that can be extended to produce all sequences as follows:

sub sumTo {
    my $target = shift;
    return map {
        $_[ $_ ] == $target
        ? [ $target ]
        : do{
            my $x = $_[ $_ ];
            map {
               @{ $_ }
               ? [ $x, @{ $_ } ]
               : ()
            } sumTo( $target - $x, @_[ 0 .. $_-1, $_+1 .. $#_ ] );
        };
    } 0 .. $#_;
}

my @numbers = ( 5, -6, 8, 10, 12, 3, 10 );

print "$ARGV[ 0 ]: sum [@$_] => ", sum( @$_ )
    for sumTo( $ARGV[ 0 ], @numbers );

__END__

[ 9:46:47.20] c:\test>688357 21
21: sum [5 -6 10 12] => 21
21: sum [5 -6 12 10] => 21
21: sum [5 -6 12 10] => 21
21: sum [5 -6 10 12] => 21
21: sum [5 10 -6 12] => 21
21: sum [5 10 12 -6] => 21
21: sum [5 12 -6 10] => 21
21: sum [5 12 -6 10] => 21
21: sum [5 12 10 -6] => 21
21: sum [5 12 10 -6] => 21
21: sum [5 10 -6 12] => 21
21: sum [5 10 12 -6] => 21
21: sum [-6 5 10 12] => 21
21: sum [-6 5 12 10] => 21
21: sum [-6 5 12 10] => 21
21: sum [-6 5 10 12] => 21
21: sum [-6 10 5 12] => 21
21: sum [-6 10 12 5] => 21
21: sum [-6 12 5 10] => 21
21: sum [-6 12 5 10] => 21
21: sum [-6 12 10 5] => 21
21: sum [-6 12 10 5] => 21
21: sum [-6 10 5 12] => 21
21: sum [-6 10 12 5] => 21
21: sum [8 10 3] => 21
21: sum [8 3 10] => 21
21: sum [8 3 10] => 21
21: sum [8 10 3] => 21
21: sum [10 5 -6 12] => 21
21: sum [10 5 12 -6] => 21
21: sum [10 -6 5 12] => 21
21: sum [10 -6 12 5] => 21
21: sum [10 8 3] => 21
21: sum [10 12 5 -6] => 21
21: sum [10 12 -6 5] => 21
21: sum [10 3 8] => 21
21: sum [12 5 -6 10] => 21
21: sum [12 5 -6 10] => 21
21: sum [12 5 10 -6] => 21
21: sum [12 5 10 -6] => 21
21: sum [12 -6 5 10] => 21
21: sum [12 -6 5 10] => 21
21: sum [12 -6 10 5] => 21
21: sum [12 -6 10 5] => 21
21: sum [12 10 5 -6] => 21
21: sum [12 10 -6 5] => 21
21: sum [12 10 5 -6] => 21
21: sum [12 10 -6 5] => 21
21: sum [3 8 10] => 21
21: sum [3 8 10] => 21
21: sum [3 10 8] => 21
21: sum [3 10 8] => 21
21: sum [10 5 -6 12] => 21
21: sum [10 5 12 -6] => 21
21: sum [10 -6 5 12] => 21
21: sum [10 -6 12 5] => 21
21: sum [10 8 3] => 21
21: sum [10 12 5 -6] => 21
21: sum [10 12 -6 5] => 21
21: sum [10 3 8] => 21
[download]

Each is a logical progression of the preceeding.

I was just exploring what felt like a quasi-contradiction in your argument. You seemed to be arguing that the solution of interest would be the first non-empty set, and I had trouble seeing where you'd want both the "non-empty" and the "first" qualifiers.
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All, except the empty set case. Eventually you might need to investigate further to see if one such set was consistant with other information--order and delivery notes etc.--but initially, just whether any such (non-empty) set existed might be a useful boolean. But it would be rendered meaningless if the empty set is considered a match.

In truth, it was just the first example I thought of that supported my argument. However, hard as I might try, I cannot think of a counter example. Not one.