Ignore most of this...

I wouldn't spend the time or money to build a PCB for this. What you can learn from the circuit will be accomplished in a few minutes. That is just an opinion, choose to ignore. I would just breadboard it and maybe play around measuring volltages and currents with my DMM. Then move on...

Using a pot to feed the LED is a recipe for disaster. When you turn the knob toward the top of the drawing, the resistance declines and the LED current begins to increase rapidly. Unfortunately, an LED is not a linear load and once you start to flow current, it increases rapidly with increasing voltage. You can overcome this by putting a fixed 100 Ohm resistor in series with each pot. The value is too high, 50 Ohms is probably more realistic (given a 3V source), if you had but one potentiometer. Since you can have two pots in parallel both turned to 0 Ohms, you can count on the parallel 100 Ohm resistors to protect the LED.

If you really want to simplify the project, calculate the resistor required to get 10 mA through the LED (dim) and use 2 of them in parallel to get 'bright' at 20 mA.

Assuming an LED with a V

_{f} of 2V at 20 mA and a source voltage of 3V, you calculate the resistor parallel resistance as (3V - 2V) / 0.02A = 50 Ohms. This is the parallel combination so each resistor would be 100 Ohms.

This would give you dim (10 mA) when the first switch is closed and bright (20 mA) when both switches are closed.

https://www.alliedelec.com/m/d/6355b8aba0b01578df0bb7b871ceefd7.pdf