If `a+b+c!=0` and `|a b c b c a c a b|=0` , then prove that `a=b=c dot`

Answer

Proved

Explanation

$$\Delta=\left|\begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{b} & \mathrm{c} & \mathrm{a} \\ \mathrm{c} & \mathrm{a} & \mathrm{b}\end{array}\right|=0$$
$$\Rightarrow \mathrm{C}_{1}=\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$$
$$\Rightarrow \Delta=\left|\begin{array}{lll}a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b\end{array}\right|$$
$$\Rightarrow(a+b+c)\left|\begin{array}{ccc}1 & b & c \\ 1 & c & a \\ 1 & a & b\end{array}\right|=0$$

$$\Rightarrow R_{2}=R_{2}-R_{1} ; R_{3}=R_{3}-R_{1}$$
$$\Rightarrow(a+b+c)\left|\begin{array}{ccc}1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c\end{array}\right|=0$$
$$ \text{expanding along} C_{1}$$
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