Can I reduce the current/power consumption of an electrical heater by adding a resistor?
Let's explore what can be done with series combinations. We'll begin by working out the resistance of each of the elements. V=IR and P=VI, so P=V^2/R and thus R=V^2/P. The 800W element is 72 ohms, the 1200W is 48 ohms, and the 2000W is 28.8 ohms.

First, to directly answer your question, could a resistor be added in series with that 2000W element to make it heat less? Suppose we wanted to get 1500W from the combination of the element and the resistor. We'd need the combined resistance to be 240^2/1500=38.4 ohms, and the 2000W element contributes 28.8 of this, so another 9.6 ohms are needed. How much power would that new resistor need to safely dissipate? Well, the 1500W power at 240V means a current of 1500/240=6.25 amps. Then the power in that new 9.6 ohm resistor is I^2*R or 375W. It's not impossible to find a resistor capable of handling that power, but it's not exactly easy either.

OK, what would happen if two of your heaters were wired in series? Suppose we pick the 2000W and the 800W. The total resistance is 100.8 ohms and the power would be 571W. Added to the full-power 1200W unit, you'll have 1771W of heating. If the 2000W and the 1200W are in series then it's 76.8 ohms and 750W, for a room total of 1550W. Finally, if the 2000W is allowed to run full power while the 800W and 1200W units are in series, they'll have a resistance of 120 ohms and power 480W. The room total would be 2480W.

One other option.. you could consider powering any of the heaters from 120V instead of 240V. It'll deliver 1/4 of the nameplate power if you do that.