Voltage regulator too hot to touch
A linear regulator dissipates heat proportional to the amount of voltage it must drop, and the amount of current flowing through it. 

Input supply = 12 Volts

Option A: 

Output Vout = 3.3 Volts
Vfred = 2.1 Volts
Current = (Vout - Vf) / R = 7.5 mA per LED = 60 mA total
Power dissipated by regulator, P = (Vinput - Vout) x I = 522 mW

Over half a watt is being dissipated by the linear regulator in this case. Add the green and blue channels, and dissipation will tend to thrice that figure.

Option B:

Output Vout = 5 Volts
Vfred = 2.1 Volts
Current = (Vout - Vf) / R 
 = 19.33 mA per LED = 154.67 mA total
Power dissipated by regulator, P = (Vinput - Vout) x I = 1082.67 mW

Over 1 watt is being dissipated by the linear regulator in this case. 1 Watt is a substantial amount of power being emitted as heat. The regulator getting too hot to touch is not unexpected in this case. Adding the green and blue channels will make things much worse.

Recommendations: 


Use the breadboard power supply in 3.3 Volt mode until you have a good heat-sink attached, even 7.5 mA per LED channel should provide a reasonable amount of illumination from the LEDs. 
Switch over to a DC-DC switching regulator or buck regulator, those waste much less power as heat.