$$
\begin{aligned}
&\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|=\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
b & c & a \\
c & a & b
\end{array}\right| \\
&=(a+b+c)\left|\begin{array}{lll}
1 & 1 & 1 \\
b & c & a \\
c & a & b
\end{array}\right| \\
&=(a+b+c)\left|\begin{array}{ccc}
0 & 0 & 1 \\
b-c & c-a & a \\
c-a & a-b & b
\end{array}\right| \\
&=(a+b+c)\left[a b+b c+c a-a^2-b^2-c^2\right] \\
&=-(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]
\end{aligned}
$$
Since $a, b, c$ are sides of $a$ scalene triangle, therefore at least two of the $a, b, c$ will be unequal.
$\begin{array}{ll}\therefore \quad & (a-b)^2+(b-c)^2+(c-a)^2>0 \\ & \text { Also } a+b+c>0 \\ \therefore \quad & -(a+b+c)\left[(a-b)^2+\right. \\ & \left.(b-c)^2+(c-a)^2\right] < 0\end{array}$