Explanation

(c)
Let the first term and common difference of an AP are $A$ and $D$ respectively.
$\therefore a = A + (p - 1)D,\ b = A + (q - 1)D,$
and $c = A + (r - 1)D$
Now, $\left| \begin{matrix}
a & p & 1 \\
b & q & 1 \\
c & r & 1
\end{matrix} \right| = \left| \begin{matrix}
A + (p - 1)D & p & 1 \\
A + (q - 1)D & q & 1 \\
A + (r - 1)D & r & 1
\end{matrix} \right|$
Applying $C_{1} \rightarrow C_{1} - DC_{2} + DC_{3}$
$= \left| \begin{matrix}
A & p & 1 \\
A & q & 1 \\
A & r & 1
\end{matrix} \right| = A\left| \begin{matrix}
1 & p & 1 \\
1 & q & 1 \\
1 & r & 1
\end{matrix} \right| = 0$ $(\because$ two columns are identical)