Electronic – Why are resistors used in a Zener clamp circuit

This behavior is expected based on your component selection. All zener diodes leak, but typically that leakage current is negligible. Since you measured 2.8V at point B, then the resistor is experiencing a 2.2V drop. The leakage through the diode is getting its current from node B1-7, so it's passing through the 100k resistor first, creating the 2.2V drop. Let's use Ohm's Law to determine the apparent leakage current: $$I=\frac{V}{R}=\frac{2.2V}{100k\Omega}=22{\mu}A$$ So the zener diode appears to be leaking 22uA. Totally believable. Essentially, the resistor and diode are together acting like a voltage divider.

The reason this happened was because your resistor value is very large. When the current in the circuit is in the same order of magnitude as leakage currents, strange behaviors like this will appear. Suddenly those negligible currents aren't so negligible compared to the signal current.

To avoid this problem, either choose a zener diode with a much lower leakage current (the hard way) or choose a smaller valued resistor (the easy way). This problem would go away if you used, say, a 1k resistor instead (i.e., 22uA through a 1k resistor would drop 22mV instead of 2.2V).

The zeners look like they're clamping the Gate-Source voltage to prevent damage to the output transistors. Looking at the datasheet for the 2SK134, the MAX Vgs = ±14V, and the Zeners are 12V, add in the forward voltage drop across the series diode, and it probably limits the Vgs to just under 13V.

It looks like if you happen to go over that 13V the zeners will save the output transistors and the feedback should limit the signal in the preamp stage.