Explanation

(c)
Since, the given lines are concurrent.
$\therefore\left| \begin{matrix}
a & k & 10 \\
b & k + 1 & 10 \\
c & k + 2 & 10
\end{matrix} \right| = 0 \Rightarrow 10\left| \begin{matrix}
a & k & 1 \\
b & k + 1 & 1 \\
c & k + 2 & 1
\end{matrix} \right| = 0$
Applying $R_{2} \rightarrow R_{2} - R_{1}\ and\ R_{3} \rightarrow R_{3} - R_{1}$
$\Rightarrow \ 10\left| \begin{matrix}
a & k & 1 \\
b - a & 1 & 0 \\
c - a & 2 & 0
\end{matrix} \right| = 0$
$\Rightarrow \ 10\left\lbrack 1(2b - 2a - c + a) \right\rbrack = 0$
$\Rightarrow \ 2b = a + c$
Hence, $a,b$ and $c$ are in AP