If a, b, c are sides of a scalene triangle, then the value of | … | is :

$$\begin{aligned} &\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|=\left|\begin{array}{ccc} a+b+c & a+b+c & a+b+c \\ b & c & a \\ c & a & b \end{array}\right| \\ &=(a+b+c)\left|\begin{array}{lll} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{array}\right| \\ &=(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ b-c & c-a & a \\ c-a & a-b & b \end{array}\right| \\ &=(a+b+c)\left[a b+b c+c a-a^2-b^2-c^2\right] \\ &=-(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \end{aligned}$$​​abc​bca​cab​​=​a+b+cbc​a+b+cca​a+b+cab​​=(a+b+c)​1bc​1ca​1ab​​=(a+b+c)​0b−cc−a​0c−aa−b​1ab​​=(a+b+c)[ab+bc+ca−a2−b2−c2]=−(a+b+c)[(a−b)2+(b−c)2+(c−a)2]​ Since $a, b, c$a,b,c are sides of $a$a scalene triangle, therefore at least two of the $a, b, c$a,b,c will be unequal. $\begin{array}{ll}\therefore \quad & (a-b)^2+(b-c)^2+(c-a)^2>0 \\ & \text { Also } a+b+c>0 \\ \therefore \quad & -(a+b+c)\left[(a-b)^2+\right. \\ & \left.(b-c)^2+(c-a)^2\right] < 0\end{array}$∴∴​(a−b)2+(b−c)2+(c−a)2>0 Also a+b+c>0−(a+b+c)[(a−b)2+(b−c)2+(c−a)2]<0​