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