Explanation

(c)
Now, radical axis of circles $S_{1}$ and $S_{2}$ is
$S_{1} - S_{2} = 0$
$\Rightarrow \ \ \ \ x^{2} + y^{2} - 6x - 6y + 4 - x^{2} - y^{2} + 2x + 4y - 3 = 0$
$\Rightarrow \ \ 4x + 2y - 1 = 0$ …(i)
Radical axis of circle $S_{2}$ and $S_{3}$ is
$S_{2} - S_{3} = 0$
$\Rightarrow \ \ \ x^{2} + y^{2} - 2x - 4y + 3 - x^{2} - y^{2} - 2ky - 2y - 1 = 0$
$\Rightarrow \ \ \ \ (2 + 2k)x + 6y - 2 = 0$ …(ii)
For existence of radical centre
$\left| \begin{matrix}
4 & 2 \\
2 + 2k & 6
\end{matrix} \right| \neq 0$
$\Rightarrow \ 24 - 2(2 + 2k) \neq 0\ \ \Rightarrow \ \ k \neq 5$