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

Where,

x_{1} = 2a, y_{1} = 2, z_{1} = 6;

x_{2} = -4, y_{2} = 3b, z_{2} = -10;

x_{3} = 8, y_{3} = 14, z_{3} = 2c

We know that the coordinates of the centroid of the triangle, whose vertices are

(x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}), are

\( \dfrac{(x_1+x_2+x_3)}{3},\dfrac{(y_1+y_2+y_3)}{3},\)

\( \dfrac{(z_1+z_2+z_3)}{3}\)

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

For triangle PQR , the coordinates will be,

ΔPQR = \( \dfrac{2a-4+8}{3}=\dfrac{2+3b+14}{3}=\dfrac{6-10+2c}{3}\)

ΔPQR = \( \dfrac{2a+4}{3}=\dfrac{3b+16}{3}=\dfrac{2c-4}{3}\)

Now, we are given that centroid is the origin,

2a + 4 = 0, 3b + 16 = 0, 2c – 4 = 0

a = -2, b = \( \dfrac{-16}{3}\), c = 2

The values of a, b and c are a = -2,

b = \( -\dfrac{16}{3}\), c = 2

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