f(x)=ax2+bx+c
f(5)=25a+5b+c
f(2)=4a+2b+c
25a+5b+c = -3(4a+2b+c)
25a+5b+c = -12a -6b -3c
37a+11b+4c=0
f(5)=-3f(2)
3 is a root, hence
f(3)=0,
9a+3b+c=0
37a+11b+4c=0 -------(i)
9a+3b+c=0 ----------(ii)
On solving (i) and (ii) we get, b = a, c = -12a -----(iii)
Substituting (iii) in f(x) = 0
ax2 + ax – 12a = 0 => a(x2 + x – 12) = 0
Given that a ≠ 0 =>(x-3)(x+4) = 0 => x = 3 (or) -4
x=3 root is already given hence the other root is -4
but With the given two conditions (i) and (ii), it is not possible to find the value of a+b+c