R=V/I but that alone is not very useful without knowing the circuit. If the full voltage (12V) is acros the resistor, that equation can be used directly. However, if there are any other circuit elements in the current path their current-voltage characteristic has to be taken into account.

If you use the 2N2222 just as a switch, then the voltage drop between collector and emitter is ~1 V as long as the transistor is operated in saturation (base current >= Ic/40). This voltage drop is not available at the resistor, so the voltage across the resistor is only 11 V.

Beware of the power loss in the transistor. At Vce=1V and Ice=0.7A, the total power converted to heat in the transistor is at least 0.7W (not counting the contribution by the base current). As the 2N2222 in TO-92 case has a thermal resistance of 200 °C/W (Junction ambient), this power will heat up the junction to 0.7*200 °C = 140 °C above ambient temperature. Add 25 °C for ambient temperature and the junction will be hot at 165 °C. Max. junction temperature is 150 °C, so you will need a heatsink.

Calculate backwards:

150 ° (max) - 25 ° (ambient) = 125 ° max. temperature drop across the thermal resistance junction-ambient. Since the case has an irreducible thermal resistance of ~84 °/W junction-case, there is a temperature gradient of 84 °/W * 0.7 W = 59 °C acros the case. So for the heatsink the remaining temperature difference is 125 ° (from above) - 84 ° = 41 °. SO the heatsink's thermal resistance must be <= 41 °/0.7W = 59 °/W.

Note all temperatures in ° C!

Harald