The trick is to ensure that each tube type sees the correct voltage for its heater. Making the trick work requires using a power supply voltage that is a multiple of 6.3 volts. 12.6 volts would not work as any voltage drop across the series resistor would prevent the 12SN7 from being used. This makes 18.9 volts the next choice.
      Before calculating the resistor value, let's just look at the voltage ratios between tube types. Since the 6SN7 (600 mA) draws twice the heater current that a 12SN7 (300 mA) does, the resistor will experience twice the voltage drop with the 6SN7 as the 12SN7. Thus when the 12SN7 is used, the resistor will drop 6.3 volts, leaving 12.6 volts for the 12SN7. And when the 6SN7 is used, the resistor will drop 12.6 volts, leaving 6.3 volts for the 6SN7. Now lets find the resistor value. The series resistor must equal 21 ohms, as 6.3 dived by 300 mA and 12.6 divided by 600 mA both equal 21 ohms. As this resistor will dissipate 7.5 watts of heat with the 6SN7 and 3.75 watts with the 12SN7, it should be rated for at least 10 watts and 20 watts much would be safer.
    What about the 8SN7? Will it also work? Yes, but not as perfectly as we might like. The 8SN7 needs 8.4 volts at 450 mA to run its heater. Plugging it into this circuit will cause a current draw of 476 mA through the series resistor, which will cause a 10 volt voltage drop across the resistor. This will yield 8.9 volts across the 8SN7's heater. This a little high, 6 percent high, but as 8SN7s can be had for so little money that burning it out a bit faster little matters. And as the resistor will do a great deal of inrush current limiting, this circuit may prove to be ultimately easier on the heater than a fixed 8.4 volt power supply.
   This trick can be used with other tubes as well. For example, the 6BQ5 comes in two other heater versions: 8BQ5 and 10BQ5; the 6SL7 has a match in the 12SL7. But remember, only one tube per voltage dropping resistor.