Three series LED arrays in parallel, resistor value

I would not use one resistor for all LED branches. Use one resistor for each LED branch.

^{simulate this circuit – Schematic created using CircuitLab}

Uled=3V
Iled=300mA
Uledbranch=9V
Iledbranch=300mA
Vcc=12V

R= 12-9 / 3e-1 = 10 Ohms
Pr= 12-9 * 3e-1 = 0.9W

Your 12V 2A switching power supply has required current output for driving 3x300mA branches. It might flicker or get hot.

You should also try experiment with connecting 4 LEDs in series and limit the current with resistor (it will be much cheaper and smaller).

If you are connecting a resistor in series with 1 or more LEDs do this:

You will need to know the voltage drop beetwen LEDs (normally around 2V per LED) and the voltage input. The voltage that gonna drop beetwen your resistor is what didn't drop in the LEDs. So:

$$V_{in} = V_r + V_{led}$$

$$V_r = V_{in} - V_{led}$$

Then the current going thought the resistor gonna be the current you want to flow in the LEDs, most of the time I use 10mA, the max the LED can handle is normally around 20mA. That will determine how much bright your LED gonna be.

With your resistor voltage and current you can use ohms law to calculate its resistance.

$$V=R \cdot I$$

One thing to pay attention is the power dessipated in the resistor.

$$P=V_r\cdot I_r$$

If you have parallel LED association I would consider using this to calculate a resistor for each parallel. The problem with using only one resistor is that all the current from each LED will go though this resistor, so it may need to have a better power rate. Even if you get a resistor that has a good power rate you may consider calculating separated LED resistors because power rate won't prevent it from heating up.