If average of a,b,c is m and (ab+bc+ca)=0, then the average of a2,b2,c2 is :m23m29m2m

The question seems to be incorrectly formatted, making it difficult to understand what is being asked. However, I'll try to interpret it as best as I can.

Given that the average of a, b, c is m, we can write this as:

(a + b + c) / 3 = m => a + b + c = 3m -----(1)

We are also given that ab + bc + ca = 0.

We need to find the average of a^2, b^2, c^2. Let's denote this as n for simplicity, so:

(a^2 + b^2 + c^2) / 3 = n => a^2 + b^2 + c^2 = 3n -----(2)

We can express a^2 + b^2 + c^2 as (a + b + c)^2 - 2(ab + bc + ca) using the identity (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).

Substituting the given values from equation (1) and the given condition ab + bc + ca = 0, we get:

a^2 + b^2 + c^2 = (3m)^2 - 2*0 => a^2 + b^2 + c^2 = 9m^2

So, the average of a^2, b^2, c^2 is 9m^2 / 3 = 3m^2.

Therefore, the answer is 3m^2.

The given question seems to be incomplete or incorrectly formatted. However, I can provide a solution based on the information given.

Given that the average of a, b, c is m, we can write this as:

(a + b + c) / 3 = m => a + b + c = 3m -----(1)

Also given that ab + bc + ca = 0.

We need to find the average of a², b², c².

We know that (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Substituting the given values in the above equation, we get:

(3m)² = a² + b² + c² + 2*0 => 9m² = a² + b² + c²

So, the average of a², b², c² is (a² + b² + c²) / 3 = 9m² / 3 = 3m².