Explanation

(b)
We have, $\left| \begin{matrix}
1 + a & 1 & 1 \\
1 & 1 + b & 1 \\
1 & 1 & 1 + c
\end{matrix} \right| = \lambda$
Applying $C_{2} \rightarrow C_{2} - C_{1}\ and\ C_{3} \rightarrow C_{3} - C_{1}$
$\Rightarrow \ \left| \begin{matrix}
1 + a & - a & - a \\
1 & b & 0 \\
1 & 0 & c
\end{matrix} \right| = \lambda$
On expanding w.r.t. $R_{3}$, we get
$ab + bc + ca + abc = \lambda$ ...(i)
Given $a^{- 1} + \ b^{- 1} + c^{- 1} = 0$
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$
$\Rightarrow ab + bc + ca = 0$
From Eq. (i), $\lambda = abc$