Most of the pictures I have don't help because R212 is inside a fabric tube. The position of the resistor on the original board is highly distinctive, hidden UNDER the big heat sink plate, and relatively alone, next to the two 3.3k 2w resistors (r102, R308) that drop the relay coil voltages, in an uncomplicated portion of the pattern.
R212 is the -80v power supply (30mA) and then the -34v power supply (64mA). Call it 100mA current.
The WHOLE idea is that those series resistors would act as fuses in serious overload situations, they having an INTENDED 5.6 ohms and 1/2 watt specification. They are not intended nor needed for their voltage drop directly, but rather for their power handling ability and their reaction to an overload (pop).
Thus the WATTAGE changes are the WORST part of substitutions.
hmmm....
original is 5.6 ohms at 0.5 watt and yours is 22 ohms at 2 watts.
p = i*i * R so at 100ma and 5.6 = 0.056 w at 100ma and 22 ohms = .22w
but what current causes rated dissipation?
0.3 amp = square root of 0.5w / 5.6 ohms , while square root of 2w / 22 ohms is 0.3w too.
so we look at the voltage drop.
e = i * r so at 100ma and 5.6 ohms = .56 volt at 100ma and 22 ohms is 2.2 volt
hmmm.. a drop in the bucket so to speak...
Thus we have to look at the resistor types. the 5.6 is a carbon comp, what is yours. if it is a wire wound, it is FAR more tolerant of an overload, which is BAAAAD.
The detraction would seem that the 22 ohm would be contributing to the heat load, but it's only shifting it slightly away from the pass transistor...
DOES IT LOOK LIKE A FACTORY PART INSTALLATION???