Explanation

(b)
Take $a,b,c$ common from $R_{1},R_{2},R_{3}$ respectively,
$\therefore\ \mathrm{\Delta} = abc\left| \begin{matrix}
\frac{1}{a} + 1 & \frac{1}{a} & \frac{1}{a} \\
\frac{1}{b} + 1 & \frac{1}{b} + 2 & \frac{1}{b} \\
\frac{1}{c} + 1 & \frac{1}{c} + 1 & \frac{1}{c} + 3
\end{matrix} \right|$
Applying $R_{1} \rightarrow R_{1} + R_{2} + R_{3}$
$\mathrm{\Delta} = abc\left( 3 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)\left| \begin{matrix}
1 & 1 & 1 \\
1 + \frac{1}{b} & 2 + \frac{1}{b} & \frac{1}{b} \\
1 + \frac{1}{c} & 1 + \frac{1}{c} & 3 + \frac{1}{c}
\end{matrix} \right|$
Now, applying $C_{3} \rightarrow C_{3} - C_{2}$ and $C_{2} \rightarrow C_{2} - C_{1}$ and on expanding, we get
$\mathrm{\Delta} = 2abc\left\lbrack \ 3 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right\rbrack = 0$
$\because a \neq 0,\ b \neq 0,\ c \neq 0$
$\therefore a^{- 1} + b^{- 1} + c^{- 1} = - 3$