Let $\mathrm{A}=\left|\begin{array}{ccc}\mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{b} & \mathrm{c} & \mathrm{a} \\ \mathrm{c} & \mathrm{a} & \mathrm{b}\end{array}\right|$
$$
\begin{aligned}
&=\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
b & c & a \\
c & a & b
\end{array}\right|\left[\because R_1 \rightarrow R_1+R_2+R_3\right] \\
&=(a+b+c)\left|\begin{array}{ccc}
1 & 1 & 1 \\
b & c & a \\
c & a & b
\end{array}\right|=(a+b+c)\left|\begin{array}{ccc}
0 & 0 & 1 \\
b-a & c-a & a \\
c-b & a-b & b
\end{array}\right| \\
&=(a+b+c)[1(b-a)(a-b)-(c-a)(c-b)]
\end{aligned}
$$


$$
\begin{aligned}
&=\frac{-1}{2}(a+b+c) \times(-2)\left(-a^2-b^2-c^2+a b+b c+c a\right) \\
&=\frac{-1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]
\end{aligned}
$$
Also, $\mathrm{A}=0$
$$
\begin{aligned}
&\Rightarrow \frac{-1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]=0 \\
&\Rightarrow(a-b)^2+(b-c)^2+(c-a)^2=0
\end{aligned}
$$
$\Rightarrow(a-b)^2+(b-c)^2+(c-a)^2=0, \quad[\because a+b+c \neq 0$, given $]$
$\Rightarrow \quad a-b=b-c=c-a=0$ $a=b=c$