#### Need Solution for R.D.Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 49  Maths Textbook Solution.

Answer: $a+b+c=0 \& a=b=c$

Hint Use determinant formula

Given:   $\text { If } a, b, c \text { are real no. such that }\left|\begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|=0 \text { show that either } a+b=c \text { or } a=b=c$

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{lll} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| \\ &\text { Apply } R_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} 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{array}\right| \end{aligned}

\begin{aligned} &2(a+b+c) \text { common from } \mathrm{R}_{1} \\ &=2(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| \end{aligned}

\begin{aligned} &\text { Apply C }_{1} \rightarrow C_{1}-C_{2} \& \mathrm{C}_{2} \rightarrow C_{2}-C_{3} \\ &=2(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ c-b & a-c & b+c \\ a-c & b-a & c+a \end{array}\right| \end{aligned}

\begin{aligned} &\text { Expand from } \mathrm{R}_{1}\\ &=2(a+b+c)\left\{1 \mid \begin{array}{ll} c-b & a-c \\ a-c & b-a \end{array}\right\}\\ &(a+b+c)=0 \quad \text { or } \quad b c-b^{2}-a c+a b-(a-c)^{2}=0\\ &b c-b^{2}-a c+a b=a^{2}+c^{2}-2 a c\\ &a^{2}+b^{2}+c^{2}-a^{2}-a b-a c=0\\ &2\left(a^{2}+b^{2}+c^{2}\right)-2 a c-2 a b-2 b c=0\\ &(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0\\ &a-b=0, \quad b-c=0, \quad c-a=0\\ &a=b, b=c, \quad c=a \end{aligned}

Hence,$a+b+c=0 \& a=b=c$