Using a Pot/Variable resistor to control upto 8.5A probably wont be too efficient.
All a Pot does it resist current, and drops a voltage across it.
Or in other words, uses up power and gives it off as heat instead of the load getting the power.
If you still want to use a Variable Resistor, you can work out what wattage you need using some formula's.
Power (in watts) = I (current in amps) X V (voltage in volts)
&
V (voltage in volts) = I (current in amps) X R (resistance in ohms)
12 volt is a constant. And i'm gussing its DC.
(AC can get tricky with power factor, because of capacitive or inductive loads. Dont worry about it if you dont know what it is)
So 12 volts DC.
Say if you have a 10Kohm Pot.
(I = V/R)
At 10K you will have 0.0012amps flowing (providing your load is 0ohms which will never happen, so everything changes because all loads have resistance, or its not a load. So now it becomes a series resistance circuit with voltage dropped across your pot and the load.)
But anyway say for instance there is 0.0012amps flowing (1.2 milliamps),
You will need a 0.0144Watt Pot (14 milliwatt)
(P = I X V)
So with a 1/4W Pot (250 milliwatt)
Your Pot will be able to dissapate the heat no problem
BUT if your resistance drops to 400ohms say
you will have 30 milliamps flowing. (I = V / R)
Therefore at 30 milliamps your power consumption by the Pot will be 0.36W (360milliwatt)
AND this exceeds the rating of a Pot rated at 1/4W
and will cook.
simple maths will tell you what size (in wattage) your pot needs to be to successfully disapate enough heat without cooking....
or if you want set voltages, just use a voltage regulator try the 78xx series (7805, 7809 7812).
And if you want variable voltages try a LM317 variable voltage reg.
Teh datasheets will show you the circuit. just google those part numbers.
What do you want to use the switch mode power supply for?
Is it out of a computer?