If the mean of a, b, c is M and ab + bc + ca = 0, then the mean of \(\frac{a^2+b^2+c^2}3\) is :

A M

B 2M

C \(3M^2\)

D \(M^2\)

Correct Answer: Option C

We have : ( a + b + c) / 3 = M or (a + b + c) = 3M.

Now. \(\left(a+b+c\right)^2=\left(3M\right)^2=9M^2\)

\(<=>a^2+b^2+c^2+2\left(ab+bc+ca\right)=9M^2\)

\(a^2+b^2+c^2=9M^{2\;\;\;},(ab+bc+ca=0)\)

Required mean = \(\left(\frac{a^2+b^2+c^2}3\right)=\frac{9M^2}3=3M^2\)