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

(A) Given $\Delta=\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}=0$.
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$,we get:
$\Delta=\begin{vmatrix} 2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}=2(a+b+c)\begin{vmatrix} 1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}=0$.
Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$,we get:
$\Delta=2(a+b+c)\begin{vmatrix} 1 & 0 & 0 \\ c+a & b-c & b-a \\ a+b & c-a & c-b \end{vmatrix}=0$.
Expanding along $R_{1}$,we get:
$\Delta=2(a+b+c)[(b-c)(c-b)-(b-a)(c-a)] = 2(a+b+c)[-(b-c)^2 - (b-a)(c-a)] = 2(a+b+c)[-b^2-c^2+2bc - (bc-ab-ac+a^2)] = 2(a+b+c)[-a^2-b^2-c^2+ab+bc+ca] = 0$.
This implies $a+b+c=0$ or $a^2+b^2+c^2-ab-bc-ca=0$.
Multiplying by $2$,we get $2a^2+2b^2+2c^2-2ab-2bc-2ca=0$,which is $(a-b)^2+(b-c)^2+(c-a)^2=0$.
Since the sum of squares is zero,each term must be zero: $a-b=0, b-c=0, c-a=0$,which implies $a=b=c$.