Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a b c+b c+c a+a b$
Show that $\left|\begin{array}{ccc}1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c\end{array}\right|=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a b c+b c+c a+a b$
Taking out factors $a, b, c$ common from $\mathrm{R}_{1}, \mathrm{R}_{2}$ and $\mathrm{R}_{3},$ we get
$\text { L.H.S. }=a b c\left|\begin{array}{ccc}
\frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\
\frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\
\frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1
\end{array}\right|$
Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3},$ we have
$\Delta=a b c\left|\begin{array}{ccc}
1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
\frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\
\frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1
\end{array}\right|$
$=a b c\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left|\begin{array}{ccc}
1 & 1 & 1 \\
\frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\
\frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1
\end{array}\right|$
Now applying $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1},$ we get
$\Delta = \operatorname{abc} \left( {1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)\left| {\begin{array}{*{20}{c}}
1&0&0 \\
{\frac{1}{b}}&1&0 \\
{\frac{1}{c}}&0&1
\end{array}} \right|$
$ = abc\left( {1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)[1(1 - 0)]$
$ = abc\left( {1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)$
$ = abc + bc + ca + ab = {\text{R}}.{\text{H}}.{\text{S}}$
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