(d) $\Delta = \left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}\,} \right|$

= $abc\,\left| {\,\begin{array}{*{20}{c}}{\frac{1}{a} + 1}&{\frac{1}{b}}&{\frac{1}{c}}\\{\frac{1}{a}}&{\frac{1}{b} + 1}&{\frac{1}{c}}\\{\frac{1}{a}}&{\frac{1}{b}}&{\frac{1}{c} + 1}\end{array}\,} \right|$, by $\begin{array}{l}{C_1} \to \frac{1}{a}{C_1}\\{C_2} \to \frac{1}{b}{C_2}\\{C_3} \to \frac{1}{c}{C_3}\end{array}$

=$abc\,\left| {\,\begin{array}{*{20}{c}}{\left( {1 + \sum \frac{1}{a}} \right)}&{\frac{1}{b}}&{\frac{1}{c}}\\{\left( {1 + \sum \frac{1}{a}} \right)}&{\frac{1}{b} + 1}&{\frac{1}{c}}\\{\left( {1 + \sum \frac{1}{a}} \right)}&{\frac{1}{b}}&{\frac{1}{c} + 1}\end{array}\,} \right|$by

${C_1} \to {C_1} + {C_2} + {C_3}$

=$abc\,\left( {1 + \sum \frac{1}{a}} \right)\,\left| {\,\begin{array}{*{20}{c}}1&{\frac{1}{b}}&{\frac{1}{c}}\\1&{\frac{1}{b} + 1}&{\frac{1}{c}}\\1&{\frac{1}{b}}&{\frac{1}{c} + 1}\end{array}\,} \right|$

$[{\rm{By   taking }}\left( {\sum \frac{1}{a} + 1} \right)\,{\rm{as common]}}$

$\Delta = abc\,\left( {1 + \sum \frac{1}{a}} \right)\,\left| {\,\begin{array}{*{20}{c}}1&{\frac{1}{b}}&{\frac{1}{c}}\\0&1&0\\0&0&1\end{array}\,} \right|$,

by $\begin{array}{l}{R_2} \to {R_2} – {R_1}\\{R_3} \to {R_3} – {R_1}\end{array}$

= $abc\,\left( {1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)\,.\,1,\,$ (by expansion along ${C_1}$)

Therefore,$\Delta = 0$ $ \Rightarrow {\rm{Either}}\,{\rm{ }}abc = 0 $or $1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$

But $a,b,c$ are non-zero and hence the product $abc$ cannot be zero.

So the only alternative is that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = – 1$.