So you want to drive a 30mA load, but your PIC will only give you 20mA? You can use a BJT as a driver. The math is quite simple.

First consider the load: 3 super bright LED's, they want to run at 30mA and have a forward voltage drop of 3 volts. You have a +12V line you can use that will provide the current. This is a good start. The 2N3904 NPN transistor has a maximum current rating of 200mA, so it should work fine. The average gain (hFE) for the '3904 is 100, so with 20mA into the base we could drive 2000mA, which would fry the transistor. Better keep the base current under 2mA (2*100=200mA, the max current).

To get 30mA load current we would need just 0.3mA base current. Let's aim for 1.5mA. This is five times the minimum required current, and will help keep the transistor saturated if the supply voltage dips. So for 1.5mA on 5V from the PIC we need (5V/1.5ma)=3.333Kohms. Put that between the PIC output and the transistor base and it should work just fine.

Here is the math:
Collector current is the voltage across the resistor divided by the resistance (ignoring voltage across the transistor)

Ic = (V/Rl) = (12V-3V*3)/100 ohms = 30mA

We want the base current to saturate the transistor (5x minimum current), so we take

Ib = Ic/hFE * 5 = 30mA/100 * 5 = 1.5mA

Now to get the base resistor

Rb = Vpic / Ib = 5V/1.5mA = 3.33Kohms

If we use a 3.3Kohm resistor it gives about 1.52mA, way under the PIC output max of 20mA, so that works.

(response to uC meets BJT)