If $a, b, c$ are positive and unequal,show that the value of the determinant $\Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$ is negative.

(N/A) Applying $C_{1} \rightarrow C_{1} + C_{2} + C_{3}$ to the given determinant,we get:
$\Delta = (a + b + c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix}$
Applying $R_{2} \rightarrow R_{2} - R_{1}$ and $R_{3} \rightarrow R_{3} - R_{1}$,we get:
$\Delta = (a + b + c) \begin{vmatrix} 1 & b & c \\ 0 & c - b & a - c \\ 0 & a - b & b - c \end{vmatrix}$
Expanding along $C_{1}$:
$\Delta = (a + b + c) [(c - b)(b - c) - (a - c)(a - b)]$
$\Delta = (a + b + c) [-(b - c)^2 - (a^2 - ab - ac + bc)]$
$\Delta = (a + b + c) [-(b^2 - 2bc + c^2) - a^2 + ab + ac - bc]$
$\Delta = (a + b + c) [-(a^2 + b^2 + c^2 - ab - bc - ca)]$
$\Delta = -\frac{1}{2}(a + b + c) [(a - b)^2 + (b - c)^2 + (c - a)^2]$
Since $a, b, c$ are positive and unequal,$(a + b + c) > 0$ and $[(a - b)^2 + (b - c)^2 + (c - a)^2] > 0$. Thus,$\Delta < 0$.