I'm guessing you calculated thus:

R = V/I = 9/0.12 = 75 Ohms

Wrong. That answer is close to the effective "resistance" of one fan motor IF it continued to draw 0.12 A when supplied with only 9 V (and it would not). It's actually a good number NOT to use.

First, to reduce 12 VDC down to 9 V, you need the drop across the resistor to be 3 volts, not 9.

Next, the current flowing in the circuit. If I read your post right, EACH fan claims to draw 0.12 A, and there are two of them in parallel. If you are planning on one resistor in the line that is shared by the two fans, then the total current is 0.24 A when they have the full 12 VDC supplied. However, you propose to give them less voltage, so the current will be lower.

So, let's work with the total resistance and see where we get. Apparently each motor behaves as if it were a resistance of R = V/I = 12/0.12 = 100 Ohms, so two in parallel is like 50 Ohms. You want to add a second resistance of R in series with that, so the the voltage split is 9 V across the 50 ohms, and 3 volts across the R. Thus R is 16.67 ohms. Don't know where you can get one of those, but you might get one at 20 ohms. This would make its drop a little over 3 V, and the motors would get 8.57 V, not 9. If you did that, the current in the line would be I = V/R = 12/70 = 0.17 A, and each motor would be running on about 8.57 V with a current of 0.085 A. However, that all assumes that the motors continue to act like pure resistances, which is not true. At lower speeds the probably offer slightly less "resistance" and the real values in operation are harder to predict.

For resistor power rating, P = I^2*R = 0.17*0.17*20 = 0.58 Watts. A 1 Watt will do. However, 2 Watts is a good idea, considering the impact of starting currents.

You should be aware that, for a few seconds at start-up, a motor will pull a current of 2 to 3 times its running current because its impedance is very low when nearly stalled. In your proposed circuit, the effective "resistance" of the two motors in parallel during start-up is more like 16 to 25 ohms, not 50, and thus the voltage drop across the resistor you plan is greater - perhaps 6 V - leaving only 6 V drop across the motors to start them up. This is getting close to the minimum necessary to start the motors, and they might NOT start up this way. And even if they do when you first put this together, in two years when the motor bearings start to wear and their friction is higher, the motors will need MORE starting voltage. but won't get it. So you may be setting up a failure, now or in future. Consider how the mobo controls case and CPU fans when it is set to do so "automatically". Of course, its control system is more complicated than a simple resistor, but it does accomplish fan speed control by varying the voltage supplied to them (at least, that's how 3-pin fans work). If you observe closely, though, you will see that most of these systems work by applying the full 12 VDC to the fan for a few seconds at start-up. Then the control system takes over and reduces the voltage AFTER the fan is already running at full speed. A simple series resistor can't do that.

I can understand that you believe your other fans already provide a significant air flow into the case and through, and hence probably also through the PSU, so that the full speed of these two fans is not needed. But you don't know that, and you really have no way of testing whether some components inside your PSU are overheating. So doing what you plan is taking a risk with no way to check it yourself.