OK, I’m not really into this, but :
given the fact that an ext. PSU haven’t blown anything, and given the fact that the actual current (pushed by the ext. PSU)only dropped with the time powered, I would assume that nothing is wrong with the resistance on a 3,3V rail.
If P(Watts) = I2 x R (I squared times the resistance), it should make: 11.5 Amps X 11.5 Amps X 0.16 Ohms = 132.25 Amps x 0.16 Ohms = 21.16 Watts.
There are at least 4 heat-sinked IC’s dissipating (in a worst case scenario) a total of 21.16 Watts.
Divided by 4 to approximate each of the heat-sinked IC’s, that gives a 5.29 Watts per IC.
I don’t find that fact that much unusal.
Good luck with your repair !
Best regards,
Alan