You need to stop thinking about the transistor as a resistor... it acts like a diode, so best to think of it as one.
Here are your steps to analyze the circuit:
Need Ic = 10 mA (max)
Most BJT's need 0.7V across the base-emitter junction to turn on.
Looking at hFE (DC current gain) on the spec sheet, you'll see that a BC109 has a minimum hFE of 200. We know that Ic = Beta*Ib in the linear region (Beta is equivalent to hFE), so Ib needs to be at least 0.010 A/200 = 50 uA.
Now, if you know that the MCU will output 3.3V, and the base of the transistor will be 0.7V above ground (assuming the emitter is grounded), then you will have to drop 3.3-0.7 = 2.6V across a series resistor going into the base pin.
Now we know V and I for the base resistor, so we can use Ohms Law to find resistance. V=I*R... 2.6=50e-6*R, so R=52 kohms. This is the largest resistor that you should use (if you make R bigger, then the base current will decrease, which in turn decreases the collector current, which would starve the sensor of current). I would try using a 20k, 10k, or 5k resistor, to make sure that you drive the transistor into saturation (Vce ~0.2V), which makes it act like a switch (low voltage drop from collector to emitter).
As for leakage current, you shouldn't see any if your MCU can pull the base of transistor low and hold it there.