If four real numbers a, b, c, x satisfy $abc\neq 0$ and
$$\frac{xb+\left(1-x\right)c}{a}=\frac{xc+\left(1-x\right)a}{b}=\frac{xa+\left(1-x\right)b}{c}$$
then either a = b = c, or a + b + c = 0
Here is my attempt
------- is the part which can be filled by symmetry
The question is why am I getting the condition which is supposed was not true, also initially I divided by $a-b$ so $a=b$ doesn't make sense now???
I don't want a new solution to this question, I want to know why my solution is giving me weird results
You assumed that $a,$ $b$ and $c$ are different numbers and you got a contradiction.
It ends the proof!
I like the following way:
Let $$\frac{xb+\left(1-x\right)c}{a}=\frac{xc+\left(1-x\right)a}{b}=\frac{xa+\left(1-x\right)b}{c}=k.$$ Thus, $$bx-cx=ka-c$$ $$cx-ax=kb-a$$ and $$ax-bx=kc-b,$$ which gives $$0=k(a+b+c)-(a+b+c),$$ which gives $$a+b+c=0$$ or $$k=1.$$
For $k=1$ we obtain: $$(b-c)x=a-c,$$ $$(c-a)x=b-a$$ and $$(a-b)x=c-b.$$ Now, let $a\neq b$.
Thus, $c\neq a$ and from here $b\neq c$.
Thus, $$\frac{a-c}{b-c}=\frac{b-a}{c-a}=\frac{c-b}{a-b},$$ which gives $$(a-b)^2=(a-c)(c-b),$$ $$(a-c)^2=(a-b)(b-c)$$ and $$(b-c)^2=(a-b)(c-a),$$ which gives that $c$ is placed between $a$ and $b$, $b$ is placed between $a$ and $c$ and $a$ is placed between $b$ and $c$, which is impossible and we got a contradiction.
Id est, $a=b$ and we obtain: $$a=b=c.$$