Unless your supply is designed to current *limit* (by design), the 2.1 amps is a 'maximum available' number, and could even involve a fuse. The value may not be well controlled (the maker is happy, and so is the customer) if more current is safely available when it's being used as intended. In series, no light. In parallel, no dice (pun intended).
The simple (sound) way IMO is to put a correctly sized resistor in series with each to set it's current to the maximum allowable value, .35 amps. Why abuse the parts (shortening their lifetime, often severely)? It could also easily result in failure of the supply with attendant fire possibilities. A 'back of the envelope' value (six volts less 3 for the LED divided by .35 amps) says 7 ohms. One of these is already in the star (what controls current in the otherwise uncontrolled 'direct drive' circuits), so we need to add six ohms in series with each star. Power across the resistor will be roughly 3 volts times .35 amps, a Watt. Use two Watt resistors. Or two 12 ohm, one Watters (which would be smaller) in parallel for each LED and you should be fine. Leave the resistors out and all sorts of fun stuff can happen. With little to control current division, you could determine destructively the true strength of your LEDs. Weakest first, followed by the others in sequence until the strongest faces the full supply alone....... Or worse.
If it were me, I'd even check the six volts under load and be prepaired to adjust (we call it "trim" or "factory select") the resistor values (easy to do in very small steps with the two resistor's in parallel since you can change one at a time).
And like Mr. Bulk say's "always measure your current".
Or so I see it.
Cheers.
Doug Owen