It appears that the values given for resistors R1 (91 kΩ), R2 (91 kΩ), and R4 (100 kΩ - variable) are at least one order of magnitude too large. Depending on the C12 version, the current consumption of the dual triode tube could be between 1.2 mA (C12 VR) and as high as 2.4 mA (original C12). The nominal B+ voltage as present across C4||R6 should be 120V. Applying Ohm's Law will reveal why the voltage collapses as soon as the filament has reached the temperature for the tube to attempt to draw even the puny current of 1.2 or 2.4 mA: R1 + R2 + R4 will result in a minimum of 182 kΩ (max. 282 kΩ), and with the tube trying to draw 1.2 mA plus R6 contributing another 0.25 mA, the voltage drop across this string of resistors alone will trend toward 260 V, and (if there were more) even more for the original C12. When starting out with 260 V at C1, there won't be much left at the tube or biasing circuit.. Please check carefully what the values for R1, R2, and R4 really should be. Perhaps 9.1 kΩ for R1 and R2 are more appropriate values. Then, with R4 still adjustable between 0 and 100 kΩ, the power supply could be adjusted for B+ of the nominal 120 V, regardless whether for an original C12 drawing 2.4 mA or the later C12 VR with its 1.2 mA requirement.