clearvars

for tot=3:2000

  for a=1:tot/3

    for b=a:(tot-a)/2

      c=tot-a-b;

      s=[a b c];

      p=prod(s);

      good=true;

      for i=1:3

        tst=p/s(i)-s(i);

        l=round(log2(tst));

        if 2^l~=tst || l<0

          good=false;

          break

        end

      end

      if good

        disp(s)

      end

    end

  end

end

finds these combinations:

     2     2     2

     2     2     3

     3     5     7

     2     6    11

 

The last two sets have three different values, in case that's an implied requirement, while the first two have members that include equal values.