Given system of equationa be written as

$\left( {a – 1} \right)x – y – z = 0$

$ – x + \left( {b – 1} \right)y – z = 0$

$ – x – y + \left( {c – 1} \right)z = 0$

For non-trivial solution, we have

$\begin{array}{*{20}{c}}
{a – 1}&{ – 1}&{ – 1}\\
{ – 1}&{b – 1}&{ – 1}\\
{ – 1}&{ – 1}&{c – 1}
\end{array} = 0$

${R_2} \to {R_2} – {R_3}$

$\begin{array}{*{20}{c}}
{a – 1}&{ – 1}&{ – 1}\\
0&b&{ – c}\\
{ – 1}&{ – 1}&{c – 1}
\end{array} = 0$

${C_2} \to {C_2} – {C_3}$

$\begin{array}{*{20}{c}}
{a – 1}&0&{ – 1}\\
0&{b + c}&{ – c}\\
{ – 1}&{ – c}&{c – 1}
\end{array} = 0$

Apply, ${R_3} \to {R_3} – {R_1}$

$\begin{array}{*{20}{c}}
{a – 1}&0&{ – 1}\\
0&{b + c}&{ – c}\\
{ – 1}&{ – c}&c
\end{array} = 0$

$ \Rightarrow \left( {a – 1} \right)\left[ {bc + {c^2} – {c^2}} \right] – 1\left[ {a\left( {b + c} \right)} \right] = 0$

$ \Rightarrow \left( {a – 1} \right)\left[ {bc} \right] – ab – ac = 0$

$ \Rightarrow abc – bc – ab – ac = 0$

$ \Rightarrow ab + bc + ca = abc$