How do you resolve Polarity Stabilizer when Bifurcation Conduit has triggered?

Question

Bifurcation Conduit states:

If you have two or more opponents remaining, choose an additional
opponent.  For the rest of the turn each part of your machine affects
both chosen opponents.  Then remove this card from the game.

Polarity Stabilizer states:

Take the top three cards from the scrap pile.
Put two into your hand and one into your opponent's hand.

Assume I am playing with at least two other players and that Bifurcation Conduit has run before Polarity Stabilizer. How do I correctly resolve this card interaction?

The game currently has me in a situation to draw three cards, put two into my hand, and one into each opponents hand.

Answer 1

I don't think a definitive answer exists from the designer or publisher, but here is my informed opinion.

The first interpretation treats Polarity Stabilizer as you would any other contraption that affects your opponents. You would take the top 3 cards from the scrap pile, put 2 in your hand, then the remaining card to the first opponent. Then you would repeat this process for each remaining opponent (top 3, keep 2, 1 to opponent). This is obviously how you would play Radiometric Disruptor (Opponent scraps a card from their hand). And is probably the way you would play a card like Entropic Processor (if you have the cards to put cards from your hand into each opponents hand)

The other possible interpretation is that you take the top 3 cards from the scrap pile, put 2 in your hand, and choose an opponent to receive the remaining card. This interpretation hinges on the giving the card to your opponent part, not being considered a way to affect your opponent and the rules example :

For example, Entropic Processor states, "Put one card from your hand into your opponents hand and draw two parts from his parts pile." ... On the other hand, if you put one card from your hand into your opponent's hand and he doesn't have two cards in his parts pile, the effect takes place as much as possible. If your opponent has one card in his parts pile you get it.

I am personally leaning toward the latter interpretation.

Answer 2

Without a definitive clarification, I would be tempted to house-rule it as "three" being short-hand for "two plus one for each opponent this affects". In this case, I would say you should draw four cards, keep two and give one to each opponent.

As I say, this would just be the clarification I would apply without anything more concrete from an authority.