If two BJT bases are shorted, where does the voltage come from?
Assume each base has 0.7V across it when activated then, with T2 (or T1) activated, (dropping about 0.2V from collector to emitter), the current through R3 (or R2) is about (12V - 2*0.7V - 0.2)/33 Ω = 315 mA.

The BJT is a voltage driven device but the voltage is limited by the forward diode junction from base to emitter. If I assumed 1V across the base-emitter junctions, the current would only be a couple of tens of mA smaller into the bases.