Answer :

Given: The vertices of the triangle are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c).

x1 = 2a, y1 = 2, z1 = 6; x2 = -4, y2 = 3b, z2 = -10; x3 = 8, y3 = 14, z3 = 2c


We know that the coordinates of the centroid of the triangle, whose vertices are (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3), are .


So, the coordinates of the centroid of the triangle PQR are



Now, it is given that the origin (0, 0, 0) is the centroid.


So, we have



2a +4 = 0, 3b + 16 = 0, 2c 4 = 0



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