The big problem is that you didn't draw the schematic properly. A good schematic always makes the things clear:
simulate this circuit – Schematic created using CircuitLab
The current is \$I = \frac{U}{I}\$
I0 = I1+I2
I0 = (5V-2.065V)/270 = 10.87037037mA
I1 = (2.065V - 1.863V)/100 = 2.02mA
I2 = I0 - I1 = 8.85037037mA
Since LEDs drop voltage when there's current through them, and since Ohm's law states that:
$$ R = \frac{E}{I} $$
They can certainly be considered to be resistances of a particular value with a known current through them and a known voltage across them.
For instance, one of your your LEDs, with 20mA through it and 2 volts across it, will look like:
$$ R = \frac{E}{I} = \frac{2V}{0.02A} = 100 \text{ ohms}$$
Usually, though, the resistance of the LED is ignored because it's not needed to calculate the values of the ballast resistors or the transistor's base resistor.
The value of the ballast resistor is determined by:
$$ Rs = \frac{Vs - (n \ \ Vf) + Vce(sat)}{If}\text{ ohms} $$
where n is the number of LEDs, Vf is the forward voltage of one LED, Vce(sat) is the transistor's collector to emitter saturation voltage,\$If\$ is the LED forward current, and Vs is the supply voltage.
In your case that works out to:
$$ Rs = \frac{12V - (3 \times 2V) + 0.5V}{0.02A}\text{ 275 ohms} $$
The 330's you have in there will work, no problem, with the LEDs losing a little brightness.
Since there will be three series strings in parallel, the current into the transistor's collector will be 60 milliamperes. Switching transistors doing this kind of work are usually run at a forced beta of ten, which means that for 60 mA into the collector 6 mA is forced into the base.
The base-to-emitter junction of a transistor is basically a diode, so in this case it'll drop about 0.7 volts with 6 mA through it.
That means that with the Arduino supplying 5V to drive the base, about 4.3 volts of that must be dropped across a resistor with 6 mA through it so, from Ohm's law, R = E/I = 4.3V/6mA = 717 ohms. 750 ohms is a standard E24 value and will work well.
Best Answer
It will produce less light. Whether or not it is noticeably less depends on the LED.
Beyond that, you may get improved longevity as it will be dissipating less power - the current will be limited by the higher resistance.
You also end up with more tolerance to power supply ripple - higher voltages applied will not cause the LED to be driven out of spec.