comparing vs equating coefficients Consider two linear equations 
 Ax + By + C = 0

 ax + by + c = 0

Now if i proceed in the following way :
  Ax + By + C = ax + by + c    ( Since both are equal to zero )
 (A-a)x + (B-b)y + (C-c) = 0
 (A-a)x + (B-b)y + (C-c) = 0x + 0y + 0

which will lead me to A = a ; B = b ; C = C; assuming non-zero vlues for x and y which is clearly wrong. 
My question is why is the above wrong , and could you please clarify what does one mean by comparing coefficients and equating them ? 
 A: The big problem here, as I see it, is that you aren't entirely clear on what the equations say. Do they say that 


*

*$A, B, C, a, b, c$ are such that the equations are true for any $x, y, z$?

*$A, B, C, a, b, c$ are fixed (although possibly unknown) and we limit ourselves to values of $x, y, z$ such that both equations are satisfied simultaneously?

*$A, B, C, a, b, c$ are fixed (although possibly unknown) and we study the set of $x, y, z$ which satisfy the first equation and the set of $x, y, z$ which satisfy the second equation and see in what way we may compare the two?

*something else?


How to correctly treat the two equations will depend greatly on what they mean, and until that is clear you can't really put them togeher in any way. 
A: There is not contradiction in your result indeed the following 
$$  (A-a)x + (B-b)y + (C-c) = 0$$
is always true for every x and y if and only if the two lines coincides that is $A=a$, $B=b$,$C=c$.
Otherwise you can obtain x or y by one equation, sustitute in the other and find the intersection point.
A: The equation $$(3-2)x+(4-3)y+(6-5)=0 \\ \implies x + y + 1 = 0$$This equations holds true for all values of $x$ and $y$ at a certain line on 2D plane. So it the coefficients can be something other than $0$ and the equation will still hold. 
If you say that equation $$ax+by+c=0$$ holds for all values of $x$ and $y$, then the coefficients have to be zero. But if that isn't the case you can have non-zero coefficients.
