If M is known to be the center of gravity of the triangle ABC, then the vector am + BM + cm =?

Let a (a, 0), B (B, 0), C (C, y) be triangles
M（（a+b+c）/3,y/3）
Vector am = ((B + C-2A) / 3, Y / 3)
Vector BM = ((a + c-2b) / 3, Y / 3)
Vector cm = ((a + b-2c) / 3, - 2Y / 3)
Vector am + BM + cm = (0,0) = 0 vector

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