The easiest way to solve this problem is simple brute force.  Try every
possible combination of bars and plates on both sides, and stash ones
that are balanced into a set, and then order and print the set.
Recursion makes this easy.  Since we expect most such combinations not
to balance, things are fastest if we consider the bars only at the end.

   void explor(vector<int> &plates, vector<int> &bars, int at,
               int leftweight, int rightweight,
               set<int> seen) {
      if (at == plates.size()) {
         if (leftweight == rightweight)
            for (b : bars)
               seen.insert(b + leftweight + rightweight) ;
         return ;
      }
      explor(plates, bars, at+1, leftweight, rightweight) ;
      explor(plates, bars, at+1, leftweight+plates[at], rightweight) ;
      explor(plates, bars, at+1, leftweight, rightweight+plates[at]) ;
   }

This runs in time at least O(3^plates), with additional factors
depending on the implementation of set and the number of bars
but these factors are minimal since matches are rare.

Extra Credit:

A challenging problem is to come up with a set of n plates that has the
maximum number of balanced combinations for that n.