If the four points of intersection of the circles $x^2+y^2+ax+by+c=0$ and $x^2+y^2+a^{\text{'}}x+b^{\text{'}}y+c^{\text{'}}=0$ by the lines $Ax+By+C=0$ and $A^{\text{'}}x+B^{\text{'}}y+C^{\text{'}}=0$ respectively are concyclic, then $\left|\begin{array}{ccc}a-a\text{'}& b-b\text{'}& c-c\text{'}\\ A& B& C\\ A\text{'}& B\text{'}& C\text{'}\end{array}\right|=$

The given circles and given lines are

$$\begin{gathered} S_1\equiv x^2+y^2+ax+by+c=0 \\ {S}_2\equiv x^2+y^2+a\text{'}x+b\text{'}y+c\text{'}=0 \\ {L}_1\equiv Ax+By+C=0 \\ {L}_2\equiv A\text{'}x+B\text{'}y+C=0 \end{gathered}$$

Let $S_1=0$ meet $L_1=0$ at two points $P$ and $Q$ and $S_2=0$ meet $L_2=0$ at two points $R$ and $S$.

Further $P,Q,R$ and $S$ are given to be concyclic. Let the circle through them is

$x^2+y^2+2gx+2fy+\lambda =0....\left(i\right)$

Radical axis of $S_1=0$ and $S_2=0$ is

$S_1-S_2=0$

$\Rightarrow \left(a-a^{\text{'}}\right)x+\left(b-b^{\text{'}}\right)y+c-c^{\text{'}}=0$

The radical axis of $S_1=0$ and $S=0$ is $L_1=0$

or $Ax+By+C=0$

and radical axis of $S_2=0$ and $S=0$ is $L_2=0$

or

$A\text{'}x+B\text{'}y+C=0$

Since, the radical axes of any three circles taken in pairs ar concurrent. (i.e. lines Eqs. (ii), (iii) and (iv) are concurrent).

we have 

$\left|\begin{array}{ccc}a-a\text{'}& b-b\text{'}& c-c\text{'}\\ A& B& C\\ A\text{'}& B\text{'}& C\text{'}\end{array}\right|=0$

Hence, the correct answer is $0$