Correct option is B.
a + b + c = 0
\(\Rightarrow\) a + b = - c or b + c = -a or a + c = - b ...... (1)
If a + b + c = 0
Then (a + b + c)2 = 0
\(\Rightarrow\) a2 + b2 + c2 + 2(ab + bc + ac) = 0
\(\Rightarrow\) a2 + b2 + c2 = - 2(ab + bc + ac) ........ (2)
Also If a + b + c = 0
Then a3 + b3 + c3 = 3abc .......... (3)
Now, (a3 + b3 + c3) (a2 + b2 + c2)
= (a5 + b5 + c5) + (a3b2 + a3c2 + b3a2 + b3c2 + c3a2 + c3b2)
= (a5 + b5 + c5) + (a3b2 + b3a2 + b3c2 + c3b2 + a3c2 + c3a2)
= (a5 + b5 + c5) + (a2b2(a + b) + b2c2(b + c) + a2c2(a + c))
= (a5 + b5 + c5) + (- a2b2c - b2c2a - a2c2b) (from (1))
= (a5 + b5 + c5) - abc (ab + bc + ac)
\(\therefore\) a5 + b5 + c5 = (a3 + b3 + c3) (a2 + b2 + c2) + abc (ab + bc + ac)
= (a3 + b3 + c3) (a2 + b2 + c2) + \(\frac{(a^3 + b^3 + c^3)}{3} \times \frac{(a^2 + b^2 + c^2)}{-2}\)
(From (2) & (3))
\(= \frac{5}{6}\)(a3 + b3 + c3) (a2 + b2 + c2)
\(\Rightarrow\) \(\frac{a^5 + b^5 + c^5}{(a^2 + b^2 + c^2)(a^3 + b^3 + c^3)}\) \(= \frac{5}{6}\) \(= \frac{m}{n}\) (Given)
\(\therefore\) m = 5 & n = 6
\(\Rightarrow\) m + n = 5 + 6 = 11.