Switching 12V with an active low 5V signal
How about this for an idea.

With the input at 5V no current can flow through the zener (5 + 9 > 12). The output PNP transistor of held OFF by the base emitter resistors (= 4k7 + 2k2) and the output is 0. When the input is pulled down to 0V a small current will flow through the base  and the 2k2 resistor. The junction of the two resistors will be a 9V (the zener voltage) and the base will be at 11.4V (assuming a 0.6V Vbe drop).
The total current flowing through the base and zener (sink current) will be added (Kirchoff's current law). With values shown the base current will be 0.5mA and the resistor current 1.4mA giving a sink current of just under 2mA.