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),

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

Where,

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 

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

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

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