$\left| {\begin{array}{*{20}{c}}
a&b&c\\
b&c&a\\
c&a&b
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{a + b + c}&{a + b + c}&{a + b + c}\\
b&c&a\\
c&a&b
\end{array}} \right|$

             $ = \left( {a + b + c} \right)\begin{array}{*{20}{c}}
1&1&1\\
b&c&a\\
c&a&b
\end{array}$

             $ = \left( {a + b + c} \right)\begin{array}{*{20}{c}}
0&0&0\\
{b – c}&{c – a}&a\\
{c – a}&{a – b}&b
\end{array}$

$ = \left( {a + b + c} \right)\left[ {ab + bc + ca – {a^2} – {b^2} – {c^2}} \right]$

$ =  – \left( {a + b + c} \right)\left[ {{{\left( {a – b} \right)}^2} + {{\left( {b – c} \right)}^2} + {{\left( {c – a} \right)}^2}} \right]$

Since $a,b,c$  are sides of $a$ scalene triangle, therfore at least two of the $a,b,c,$ will be unqual.

$\therefore {\left( {a – b} \right)^2} + {\left( {b – c} \right)^2} + {\left( {c – a} \right)^2} > 0$ 

Also $a + b + c > 0$

$\therefore  – \left( {a + b + c} \right)\left[ {{{\left( {a – b} \right)}^2} + {{\left( {b – c} \right)}^2} + {{\left( {c – a} \right)}^2}} \right] < 0$