If a, b and c are real numbers, and $\Delta = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{b + c}&{c + a}&{a + b}\\{c + a}&{a + b}&{b + c}\\{a + b}&{b + c}&{c + a}\end{array}} \right| = 0$, show that either $a + b + c = 0$or $a = b = c.$

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

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

we get
$\Delta = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{2(a + b + c)}&{c + a}&{a + b}\\{2(a + b + c)}&{a + b}&{b + c}\\{2(a + b + c)}&{b + c}&{c + a}\end{array}} \right|$
Taking $2(a + b + c)$ common from ${C_1}$,

we get
$\Delta = 2(a + b + c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{c + a}&{a + b}\\1&{a + b}&{b + c}\\1&{b + c}&{c + a}\end{array}} \right|$

Applying ${R_2} \to {R_2} - {R_3}$ and ${R_3} \to {R_3} - {R_1}$ ,

we get
$\Delta = 2(a + b + c)\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&{c + a}&{a + b}\\0&{a - c}&{b - a}\\0&{b - a}&{c - b}\end{array}} \right|$

Expanding along ${C_1},$

we get
$\Delta = 2\left( {a + b + c} \right)\left[ {\left( {a - c} \right)\left( {c - b} \right) - \left( {b - a} \right)\left( {b - a} \right)} \right]$

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

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

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

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

When, $\Delta = 0$
$\Rightarrow$ $- (a + b + c)[{(a - b)^2} + {(b - c)^2} + {(c - a)^2}] = 0$

$\Rightarrow$ $a + b + c = 0$ or ${(a - b)^2},{(b - c)^2},{(c - a)^2} = 0$

$\Rightarrow$ $a + b + c = 0$ or $a = b,b = c,c = a$

$\Rightarrow$ $a + b + c = 0$ or $a = b = c$