When α = 120, that is, BM and BA are on the same straight line, point P is on the line segment BA, and (here it should be) PC > 3 √ 3.

The rest, you are right, the distance from point C to BM is the shortest, but the PC value is the smallest when it is vertical.

It's a bit far-fetched for this reason. We want to list the analytical formula PC = √( PN the square of PN+the square of NC), which is very complicated and is the knowledge of high school.

There is no need for the subject to entangle this problem, after all, it is beyond the scope.

If there is such a problem in the exam, you can write clearly that when BM is perpendicular to AC, the value of PC is the smallest. The obvious conclusion can omit the argument of superclass.