If the origin is the centroid of the triangle $P Q R$ with vertices $P (2 a, 2, 6), Q (- 4, 3 b, - 10)$ and $R (8, 14, 2 c),$ then find the values of $a, b$ and $c .$

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Solution

$P (2 a, 2, 6), Q (- 4, 3 b, - 10), Q (8, 14, 2 c) (0, 0, 0) = (\frac{2 a - 4 + 8}{3}, \frac{2 + 3 b + 14}{3}, \frac{6 - 10 + 2 x}{3}) \Rightarrow a = - 2, b = - \frac{16}{3}, c = 2$

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If the origin is the centriod of the triangle PQR with vertices $P (2 a, 2, 6), Q (- 4, 3 b, - 10)$ and $R (8, 14, 2 c),$ then find the values of a, b, and c.

If the origin is the centroid of the triangle with vertices P(2a, 2, 6), Q(-4, 3b, -10) and R(8, 14, 2c), find the values of a,b and c. Also, determine the value of $a^{2} + b^{2} - c^{2}$ .

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If the origin is the centroid of the triangle PQR with vertices P(2a,2,6),Q(−4,3b,−10) and R(8,14,2c), then find the values of a,b and c.

P(2a,2,6),Q(−4,3b,−10),Q(8,14,2c)(0,0,0)=(2a−4+83,2+3b+143,6−10+2x3)⇒a=−2,b=−163,c=2

If the origin is the centroid of the triangle $P Q R$ with vertices $P (2 a, 2, 6), Q (- 4, 3 b, - 10)$ and $R (8, 14, 2 c),$ then find the values of $a, b$ and $c .$

$P (2 a, 2, 6), Q (- 4, 3 b, - 10), Q (8, 14, 2 c) (0, 0, 0) = (\frac{2 a - 4 + 8}{3}, \frac{2 + 3 b + 14}{3}, \frac{6 - 10 + 2 x}{3}) \Rightarrow a = - 2, b = - \frac{16}{3}, c = 2$