The drop across resistors R1 and R2 is 6.3 V (12 –5 –0.7). This means that the current is 6.3mA and this flows through diode D1 when all four voltages are present. However, if for instance, the –5 V line fails, transistor T3 remains on but the base-emitter junction of T2 is no longer biased, so that this transistor is cut off. When this happens, there is no current through D which then goes out.
http://www.extremecircuits.net/2010/08/supply-voltage-monitor.html