Not sure if thats a standard transistor or a MOSFET or some other complex beastie. But if its a normal transistor: then the answer is - a DEAD transistor. The transistor is a switch. Looking at your diagram: there are pins: the base (middle left), collector (top right), and emitter (bottom right). If the base is less than 0.6v above the emitter then the switch (between collector and emitter is open). So if you tied the base to 0V then the switch would be open: collector=12v, emitter=0v. But you have connected the base to 12v which is much greater than 0.6v above the voltage on the emitter (which is 0v). This, in itself, would probably fry the device. If it doesn't then the switch will close and so you now have a short circuit between 12v and 0v and so the 'infinite' current flowing thru the transistor will kill it anyway. So you've probably killed it in two different ways !!!
That apart: the output will be 0v irrespective of what you do to the base. Once the transistor has died it will probably still be at 0v.
You need to have some sort of resitance between either: +12v and the collector, and/or the emitter and 0v, and/or the base and the input signal that is driving it.
But what are you trying to do?