Transistor switch with 2N2222A and cooling fan as load
If the switch were off, then the voltage at the collector of the transistor would be 12 volts. The fact that it's lower than that means that the transistor is conducting, which is bringing the voltage on the near side of the fan down lower than 12.

The way an NPN BJT transistor works is that the current flowing from the base to the emitter will allow a proportionally higher current too flow from the collector to the emitter. What you must figure out is how much current is flowing from the base to the emitter. If that is insufficient to cause sufficient collector current, then you must lower the resistor value.

What you likely want is to get the transistor into saturation. This means that there is insufficient voltage (because of the impedance of the load - the fan) to drive more current through the transistor. Increasing the base current won't make more collector current flow in this situation and the transistor will (more or less) act like a closed switch.

That said, if your input is a 5 volt logic high, then you're sending 33 mA (or so - the base-emitter junction won't be zero volts, after all), which "ought" to be enough to saturate that transistor.

I wonder if maybe 33 mA is too much current for the microcontroller pin in question. If so, you potentially might have (or might soon) blow that pin. You might consider increasing that resistor value. 220Ω or 330Ω are more like values I'd use under similar circumstances.