I was experimenting with ways to cut power to my board and so I added an NPN transistor on the low side of an optocoupler (Q202.) I noticed my voltage regulator was oscillating from 3.3V to GND every few seconds and then the transistor caught on fire. I am very confused as to why this would cause a catch on fire. After the transistor burned up I removed it and noticed the voltage regulator was stabilized.
No base resistors to limit current. BJTs aren't like MOSFETs. The base-emitter junction on BJT transistors is a diode and you are straight up connecting a voltage source across it.
The PN in your NPN transistors is the diode of the base-emitter junction.
Why did only Q202 catch on fire and not Q203?
No two diodes are identical due to manufacturing tolerances so when in parallel the one with the lowest voltage drop will conduct and cause the others to have insufficient voltage to fully conduct thus acting like a short-circuit across the others and hogging all the current. That's why paralleling diodes isn't good practice. And semiconductors tend to fail short so the one that flamed up was probably still conducting as a short. If it failed opened I expect Q203 to have followed it.