If $a + b + c \ne 0$ and $\left| {\begin{array}{llllllllllllllllllll}a&b&c\\b&c&a\\c&a&b\end{array}} \right| = 0$, then prove that $a = b = c$.

Let $A = \left| {\begin{array}{cccccccccccccccccccc}a&b&c\\b&c&a\\c&a&b\end{array}} \right|$

$= \left| {\begin{array}{cccccccccccccccccccc}{a + b + c}&{a + b + c}&{a + b + c}\\b&c&a\\c&a&b\end{array}} \right|$

$= (a + b + c)\left| {\begin{array}{cccccccccccccccccccc}1&1&1\\b&c&a\\c&a&b\end{array}} \right|$

$= (a + b + c)\left| {\begin{array}{cccccccccccccccccccc}0&0&1\\{b - a}&{c - a}&a\\{c - b}&{a - b}&b\end{array}} \right|$

and $\left. {{C_2} \to {C_2} - {C_3}} \right]$

Expanding along ${R_1}$,
$= (a + b + c)[1(b - a)(a - b) - (c - a)(c - b)]$

$= (a + b + c)\left( {ba - {b^2} - {a^2} + ab - {c^2} + cb + ac - ab} \right)$

$= \frac{{ - 1}}{2}(a + b + c) \times ( - 2)\left( { - {a^2} - {b^2} - {c^2} + ab + bc + ca} \right)$

$= \frac{{ - 1}}{2}(a + b + c)\left[ {{a^2} + {b^2} + {c^2} - 2ab - 2bc - 2ca + {a^2} + {b^2} + {c^2}} \right]$

$= - \frac{1}{2}(a + b + c)\left[ {{a^2} + {b^2} - 2ab + {b^2} + {c^2} - 2bc + {c^2} + {a^2} - 2ac} \right]$

$= \frac{{ - 1}}{2}(a + b + c)\left[ {{{(a - b)}^2} + {{(b - c)}^2} + {{(c - a)}^2}} \right]$

Also, $A = 0$
$= \frac{{ - 1}}{2}(a + b + c)\left[ {{{(a - b)}^2} + {{(b - c)}^2} + {{(c - a)}^2}} \right] = 0$

${(a - b)^2} + {(b - c)^2} + {(c - a)^2} = 0$

, given]
$\Rightarrow$ $a - b = b - c = c - a = 0$
$a = b = c$