@above, are you sure the problem is correct? I think instead of $A,B,C,D,E,F$ lie on one circle we must have no 4 of them are concyclic.

yes, that was a huge typo, thanks, correcting it

For any point $X$ we define three functions $f_{1}(X)=AX^{2}+BX^{2}-DX^{2}-EX^{2}$, $f_{2}(X)=BX^{2}+CX^{2}-EX^{2}-FX^{2}$, $f_{3}(X)=CX^{2}+DX^{2}-FX^{2}-AX^{2}$. It's not hard to see that $f_{1},f_{2},f_{3}$ are all linear function. (Follows from linearity of PoP in degenerate circle) By using the fact that no 4 points in $A,B,C,D,E,F$ we have $P\neq P'$, $Q\neq Q'$ and $R\neq R'$ Moreover, we have that $f_{1}(P)=f_{1}(P')=0$, $f_{2}(Q)=f_{2}(Q')=0$ and $f_{3}(R)=f_{3}(R')=0$. We divide the problem into two cases. $\textbf{Case 1.}$ If $PP'$ and $QQ'$ intersects, assume at point $Y$. As $f_{1}$ and $f_{2}$ are both linear, we'll have that $f_{1}(Y)=f_{2}(Y)=0$ $$AY^{2}+BY^{2}-DY^{2}-EY^{2}=f_{1}(Y)=0=f_{2}(Y)=BY^{2}+CY^{2}-EY^{2}-FY^{2}$$Which implies $BY^{2}+CY^{2}-FY^{2}-AY^{2}=0$ Hence, $f_{3}(Y)=0$. As $f_{3}(R)=f_{3}(R')=f_{3}(Y)=0$ and $f_{3}$ is linear, it's not hard to see that if $Y$ doesn't lie on line $RR'$, we'll have that $f_{3}(T)=0$ for any point $T$ which is impossible. (Just check it yourself) Therefore, $Y$ must lies on line $RR'$. The first case is now done. $\textbf{Case 2.}$ If $PP'\parallel QQ'$. We'll prove that $QQ'\parallel RR'$, suppose otherwise, $QQ'\cap RR'=Z$. By linearity, we'll have $f_{2}(Z)=f_{3}(Z)=0$. By using the same way as the case before, we can proof that $Z$ also lies on $PP'$, contradicting the fact that $PP'\parallel QQ'$. So, $QQ'\parallel RR'$. And the problem is now done.