Your specifications for the load weren't clear enough, so let's talk about your voltage divider first. If you want a +12V, a -12V and a +5V output, you need three resistors instead of two, with the sum of the resistance values of R1 & R2 being equivalent to the resistance value of R3. Your +12Vdrop is the sum of individual drops across R1 & R2 in series, your -12Vdrop is the drop across R3 with the ground before it, while your 5Vdrop will be the drop across R1 alone.
Your schematic should look like this:
(I used kΩ resistors to minimize power consumption through the divider although you can use resistors of smaller value as long as they are proportional this way).
The total resistance value of the entire circuit is 7.2kΩ, while supply voltage is 24V. You can see that our +12V drop is the drop across both R1 and R2 in series, meaning both the resistors' values should have the sum of half the total resistance value. 2.1kΩ+1.5kΩ = 3.6kΩ, which is half of 7.2kΩ or the said total resistance value. This means that half of the supply voltage is the voltage drop across the two resistors, hence +12V. The drop across R3, having a value of 3.6kΩ, should also have a 12V drop across it, as its resistance value is half of the total resistance value of the entire circuit. The ground is always defined as 0V, so since R3 comes after it, there has to be a negative voltage drop across R3, hence -12V. The +5V division is provided by R1, with a value of 1.5kΩ which is 1/4.8 of the total resistance value, therefore the voltage drop across it will be 1/4.8 of the supply 24V, hence the +5V drop across it. Looking at the schematic, the +5V division is embedded within the +12V division because this, together with the 2.1kΩ resistor, provides half the total resistance value of the circuit to provide the +12V.
Again I need more clarification about the load. I need to know which voltage divider the load is connected to in order to start with that 80mW being talked about. Hope I helped.