General wiring questions: high power LEDs, relay and microcontroller
I admit the explanation about the transistor is a little misleading. I wouldn't have chosen to use the term "active low" in this scenario because you have direct  control of the GPIO pin, not direct control of the relay. The GPIO pin is active high. All they're saying is you need to turn on the transistor so it can provide a path to ground for the relay's coil. And to do so, you need to drive the GPIO pin high.

Here's what the full circuit probably looks like for each GPIO/relay pair:







simulate this circuit &ndash; Schematic created using CircuitLab

When the GPIO of the RPi is driven high (in your code), the 2N222 NPN transistor is turned on. That allows current to flow from the 5V source, through the relay coil, through the transistor, and into ground, which will activate the relay. The author's description of "active low" is actually referring to the spot I marked "V1" in the circuit. When the voltage at that node is low (when the transistor is on or if you just touch a ground wire there), current will flow and the relay will activate. I think she explained it that way because that point in the circuit is physically manifested as a pin you can connect to.

R1 is a current-limiter resistor. It prevents too much current from being sourced from the GPIO pin and into the transistor's base. R2 is a pull-down resistor. It forces the base of the transistor low (and thus the transistor off) if the GPIO pin is physically disconnected or high impedance (tri-stated). When the GPIO pin is attached, R2 can be ignored. There is no signal flipping going on.

If you wire up your LEDs and current source like so, you can connect them all to the same ground wire:






simulate this circuit

I'm assuming you've got a handle on how much current these LEDs need. If you're not sure about hooking up your LEDs to a 700mA constant current source, please do more research first (or sit back and enjoy the fireworks). At no time should all of the LEDs be off - unless the current source has been safely disabled.

This circuit will work as long as you do as you said and only have one LED on at a time. If you do activate more than one at a time, the current will divide between them. And since LEDs have a negative temperature coefficient, one LED will tend to dominate the current and look brighter than the others.