# 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

**Solution:**

It is known that the coordinates of the centroid of the triangle, whose vertices are (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) are [(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3, (z₁ + z₂ + z₃)/3]

Therefore, coordinates of the centroid of

ΔPQR = [(2a - 4 + 8)/3, (2 + 3b + 14)/3, (6 - 10 + 2c)/3]

= [(2a + 4)/3, (3b + 16)/3, (2c - 4)/3]

It is given that origin (0, 0, 0) is the centroid of the ΔPQR

Hence,

(2a + 4)/3 = 0, (3b + 16)/3 = 0 and (2c - 4)/3 = 0

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

Thus, the values of a = - 2, b = - 16/3 and c = 2

NCERT Solutions Class 11 Maths Chapter 12 Exercise ME Question 3

## 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.

**Summary:**

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 the values of a, b and c are a = - 2, b = - 16/3 and c = 2