Explanation

(c)
$\left| \begin{matrix}
b^{2}c^{2} & bc & b + c \\
c^{2}a^{2}\ & ca & c + a \\
a^{2}b^{2} & ab & a + b
\end{matrix} \right|$
On multiplying $R_{1},R_{2},R_{3}$ by $a,b,c$ respectively and divide the whole by $abc$
$= \frac{1}{abc}\left| \begin{matrix}
{ab}^{2}c^{2} & abc & a(b + c) \\
{bc}^{2}a^{2}\ & bca & b(c + a) \\
a^{2}b^{2}c & abc & c(a + b)
\end{matrix} \right|$
On taking common $abc$ from $C_{1}$and$\ C_{2}$, we get
$= \frac{(abc)(abc)}{abc}\left| \begin{matrix}
bc & 1 & ab + ac \\
ca\ & 1 & bc + ab \\
ab & 1 & ca + bc
\end{matrix} \right|$
Now, $C_{1} \rightarrow C_{1} + C_{3}$
$= abc\left| \begin{matrix}
ab + bc + ca & 1 & ab + ac \\
ca + bc + ab\ & 1 & bc + ab \\
ab + bc + ca & 1 & ca + bc
\end{matrix} \right|$
$= (abc)(ab + bc + ca)\left| \begin{matrix}
1 & 1 & ab + ac \\
1\ & 1 & bc + ab \\
1 & 1 & ca + bc
\end{matrix} \right|$
$= 0\ \ \ \ \ \ \ \ \lbrack\because\$ two columns are identical]