Is this a sensbile way to switch a 12V line with 3.3V logic
Lots of problems with your circuit:

T1 is used as a emitter follower.  Its output will therefore be less than the 3.3 V input.  Figure 700 mV for the B-E drop, and the maximum the gate of Q1 is driven to is 2.6 V

Q1 is used as a source follower.  Its output will thefefore be less than its input.  Unlike a BJT, as T1 is, the G-S voltage is not so easily known.  To drive any substantial current, it must at least somewhat above the gate threshold voltage.  With only 2.6 V on the gate, there may not be any voltage left at all to drive the motor.  Basically, the motor will never be turned on by this circuit.

Even if the above weren't problems, there is nothing driving the gate of Q1 low when the motor is supposed to be off.



A simple circuit that does what you want is to use a N channel FET that can be driven well from 3.3 V as a low side switch.  For example, the IRLML6344 would be suitable here.  It has a maximum RDSON of 37 m&Omega; with only 2.5 V gate drive.

Connect the motor (with diode as you show) between the 12 V power supply and the FET drain, source to ground, and drive the gate directly from the 0 to 3.3 V digital output.  Yes, it's that easy: