The power dissipated in that resistance would be IR = 0.03 * 200 = 6W.

No it wouldn't. P = U x I = R x I x I = U x U / R

So in this case we have 0.03 x 0.03 x 200 = 0.18 W

Sanity check: the pot drops 6 volts (12-6) with 30 mA current going through it, which gives U x I = 6 x 0.03 = 0.18 W

Beware that a fairly small (depending on what value the pot is) part of the pot might be generating the heat - not the full track of the pot... Select pot size with that in mind.

/Michael