Question 481709
You have to apply the rule of identities.
 
a1/b1 = a2/b2 = a3/b3 
= (a1+a2+a3)/(b1+b2+b3) ---- 1
 = (a1-a2+a3)/(b1-b2+b3) ---- 2
 = (a2-a3+a1)/(b2-b3+b1) ---- 3
 = (a3-a1+a2)/(b3-b1+b2) ---- 4
..............
 
apply  to the given equations:
 {{{(x+y)/(3a-b) = (y+z)/(3b-c) = (z+x)/(3c-a)}}}
 = {{{(x+y+y+z+z+x) / (3a-b+3b-c+3c-a)}}} ----using (1)
 = {{{(x+y+z)/(a+b+c)}}}
 
= {{{(x+y-y-z+z+x) / (3a-b-3b+c+3c-a)}}} ---- by (2)
 = {{{2x / (2a - 4b + 4c)}}}
 = {{{x / (a-2b+2c)}}}
 = {{{ax / (a^2 - 2ab + 2ca)}}} ---- result A
 
= {{{(y+z-z-x+x+y) / (3b-c-3c+a+3a-b)}}} ---- by (3)
 = {{{2y / (2b - 4c + 4a)}}}
 = {{{y / (b-2c+2a)}}}
 = {{{by / (b^2 - 2bc + 2ab)}}} ---- result B
 
= {{{(z+x-x-y+y+z) / (3c-a-3a+b+3b-c)}}} ---- by(4)
 = {{{2z / (2c - 4a + 4b)}}}
 = {{{z / (c-2a-2b)}}}
 = {{{cz / (c^2 - 2ca + 2bc)}}} ---- result C
 
= {{{(ax + by +cz) / ( (a^2 - 2ab + 2ca) + (b^2 - 2bc + 2ab) + (c^2 - 2ca + 2bc) )}}} -------- by adding results A,B,C
 
= {{{(ax + by +cz) / (a^2 + b^2 + c^2)}}}
 
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