If a, b, c are real numbers such that

Solution:

Let $\Delta=\mid b+c c+a \quad a+b$

$c+a \quad a+b b+c$

$a+b b+c \quad c+a$

$=\mid 2(a+b+c) 2(a+b+c) 2(a+b+c)$

$\begin{array}{lll}c+a & a+b & b+c \\ a+b & b+c & c+a \mid\end{array}$             [Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$ ]

$=2(a+b+c) \mid \begin{array}{lll}1 & 1 & 1\end{array}$

$c+a+b b+c$

$a+b b+c c+a$

$=2(a+b+c) \mid \quad 1 \quad 0 \quad 0$

$c+a b-c b-a$

$a+b c-a c-b \mid$           [Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$ ]

$=2(a+b+c)\left\{1\left|\begin{array}{ll}b-c & b-a \\ c-a & c-b\end{array}\right|\right\}$

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

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

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

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

But $\Delta=0$

[Given]

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

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

$\Rightarrow(\mathrm{a}+\mathrm{b}+\mathrm{c})=0$ or $a=\mathrm{b}=\mathrm{c}$

Hence proved.