in the given regular triangle ABC find the locus of the points M inside this triangle such that MAB +MBC+MCA=pi/2
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)
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
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