The working is very obvious, if any input is high (-0.8V) then the corresponding transistor is turned on and $Q_3$ is turned off.

This causes a voltage of -1.6V to appear at emitters of both of the transistors $Q_1$ & $Q_2$ Since $V_{BB}$ = -1.3$v_1$ there is only a drop of 0.3V at $V_{BE}$ of $Q_5$ , hence it is in cut–off. Current in the 220 Ω resistance flows through the conducting transistor and the output of $Q_6$ is low.

Now if both inputs are low, only $Q_3$ will conduct, and $Q_1$ & $Q_2$ will be cut off this is because emitters will be at -2.1 V and each transistor has its base at -1.8V hence drop is only 0.3V hence they are in cut-off.

In this situation current through the 220 Ω resistance flows out through $Q_6$ and O/P is high.