Looks like that "the incenter" (i.e., the center of the inscribed circle in a traingle) is one such point.

Technically, one can write a short code where point M is randomly selected inside the triangle ABC, and the sum of the angles is computed

S = MAB +MBC+MCA.

Run the code millions of times and visualize the set of points M when S is equal to pi/2, or near equal to pi/2.

The visualization will show what can be the set (the locus) of such points M (after the code is running / simulating millions of such M's).

Hopefully, this helps.

(btw, apparently, another classical M point is the circumcenter)

What about a sharp inequality for the angle sum MAB +MBC+MCA ?

I believe we can not find any such point inside the triangle which satisfies your condition.

Each point M inside the triangle will satisfy MAB+MBA+MAC=2*pi

Angle MAB is not the same as AMB.

That’s a catch.

And regular triangle mean equilateral triangle, right?