A^2 - bc = 3 b^2 - ca = 4 c^2 - ab = 5 then value of a+b+c = ?

options : 0 1/2 1 -1?

solution
a^2 - bc = 3 ..1
b^2 - ca = 4 ...2
c^2 - ab = 5 ..3

subtract (2) from (1)

a^2-b^2 +c (a-b) = -1

or (a+b+c)(a-b) = - 1

subtract (3) from (2)

(a+b+c)(b-c) = - 1

so a- b = b-c or a,b,c are in AP

now let a= b- t and c= b+ t
from 2nd equation

b^2 - (b^2 - t^2) = 4

or t = +/- 2

so a-b = 2 or -2

so (a+b+c) = 1/2 or - 1/2

as 1/2 is a choice and - 1/2 is not a choice so ans is 1/2

this is subject to the condition that equations are consistent.

else we need to evaluate a,b,c and then the sum