Consider the three linear equations
$ax+by+c=0$
$bx+cy+a=0$
$cx+ay+b=0$
where $a,b,c\in\mathbb{R}$
Prove that
$(A)$If $a+b+c=0$ and $a^2+b^2+c^2\neq ab+bc+ac$,then prove that the lines are concurrent.
$(B)$If $a+b+c=0$ and $a^2+b^2+c^2= ab+bc+ac$,then entire $xy$ plane.
$(C)$If $a+b+c\neq0$ and $a^2+b^2+c^2\neq ab+bc+ac$,then prove that the lines are neither coincident nor concurrent.
$(D)$If $a+b+c\neq 0$ and $a^2+b^2+c^2= ab+bc+ac$,then prove that the lines are coincident.
I understood that if the lines are passes through $(1,1)$,then $a+b+c=0$ otherwise not,but i am not clear what does $a^2+b^2+c^2=ab+bc+ca$ mean?
Please help me.
If $a+b+c=0$ and $a^2+b^2+c^2=ab+bc+ac$, then it follows that $0=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$, or $a^2+b^2+c^2=-2(ab+bc+ac)$. Put this together and we will see that in fact $a^2+b^2+c^2=0$, which is only possible if a, b, and с are all 0. This would make all three equations identical and trivially correct on the entire xy plane, just like the option (B) suggested.
Upd. Sure enough, the rest is as simple as this. See, $(a-b)^2=a^2+b^2+2ab\ge0\Rightarrow a^2+b^2\ge2ab$ (equals only if $a=b$). Likewise, $$b^2+c^2\ge2bc\\ a^2+c^2\ge2ac$$ Sum all that together, and we'll get $$2a^2+2b^2+2c^2\ge2ab+2bc+2ac$$ So the identity $a^2+b^2+c^2=ab+bc+ac$ is possible only when $a=b=c$, in which case our lines are all the same (or coincident, if you want to say it the fancy way).
Is this the only case when they coincide? Well, for the two lines to coincide, their equations must be proportional to each other, as in $$ax+by+c=0\\ kax+kby+kc=0$$ Plugging our second line into it, we see that $b=ka,\;c=kb,\;a=kc$, hence $k^3=1$. If we are not going to delve into complex numbers (and trust me, we are not), then the only solution is $k=1$, which means $a=b=c$.