hello sayanta a much better way to solve is:
let (x+y) / (3a-b) = (y+z) / (3b-c) = (z+x) / (3c-a) = k
by componendo and dividendo, we have
k= [(x+y) + (y+z) + (z+x)]/[(3a – b) + (3b – c) + (3c – a)]
or k= (x+y+z)/(a+b+c).......(*)
also x+y= k(3a – b)....(1)
y+z= k(3b – c)......(2)
and z+x= k(3c – a).......(3)
solving these 3 eqns we have
x= k(a+b+c) – k(3b – c)= k(a – 2b + 2c)
y= k(b – 2c + 2a)
and z= k(c – 2a + 2b)
so, ax+by+cz= k(a^2 – 2ab + 2ac + b^2 – 2bc + 2ab + c^2 – 2ac + 2bc)= k(a^2 + b^2 + c^2)
or (ax+by+cx) / (a²+b²+c²) = k.......(#)
from (*) and (#), we have
k= (x+y+z) / (a+b+c) = (ax+by+cx) / (a²+b²+c²).
kindly approve :))
