If $$a, b, c$$ are sides of a scalene triangle, then the value of $$\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$$ is :

We are given that $$a$$, $$b$$, and $$c$$ are the sides of a scalene triangle, and we need to evaluate the determinant:

$$\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$$

To compute this determinant, we use the formula for a 3x3 matrix. The determinant $$D$$ is given by:

$$D = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix}$$

Now, we calculate each 2x2 determinant:

• The first minor: $$\begin{vmatrix} c & a \\ a & b \end{vmatrix} = c \cdot b - a \cdot a = bc - a^2$$
• The second minor: $$\begin{vmatrix} b & a \\ c & b \end{vmatrix} = b \cdot b - a \cdot c = b^2 - ac$$
• The third minor: $$\begin{vmatrix} b & c \\ c & a \end{vmatrix} = b \cdot a - c \cdot c = ab - c^2$$

Substituting these back into the expression for $$D$$:

$$D = a(bc - a^2) - b(b^2 - ac) + c(ab - c^2)$$

Expanding each term:

• $$a(bc - a^2) = a \cdot bc - a \cdot a^2 = abc - a^3$$
• $$-b(b^2 - ac) = -b \cdot b^2 + b \cdot ac = -b^3 + abc$$
• $$+c(ab - c^2) = c \cdot ab - c \cdot c^2 = abc - c^3$$

Combining all terms:

$$D = (abc - a^3) + (-b^3 + abc) + (abc - c^3) = -a^3 - b^3 - c^3 + abc + abc + abc = -a^3 - b^3 - c^3 + 3abc$$

So,

$$D = 3abc - (a^3 + b^3 + c^3) = - (a^3 + b^3 + c^3 - 3abc)$$

We recall the algebraic identity:

$$a^3 + b^3 + c^3 - 3abc = \frac{1}{2} (a + b + c) \left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right]$$

Substituting this identity into the expression for $$D$$:

$$D = - \left( \frac{1}{2} (a + b + c) \left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right] \right) = -\frac{1}{2} (a + b + c) \left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right]$$

Since $$a$$, $$b$$, and $$c$$ are sides of a triangle, they are positive real numbers. Therefore, $$a + b + c > 0$$.

Additionally, because the triangle is scalene, all sides are unequal: $$a \neq b$$, $$b \neq c$$, and $$c \neq a$$. This means:

• $$(a - b)^2 > 0$$
• $$(b - c)^2 > 0$$
• $$(c - a)^2 > 0$$

So, the sum $$(a - b)^2 + (b - c)^2 + (c - a)^2 > 0$$.

Thus, the product $$(a + b + c) \left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right]$$ is positive (since both factors are positive).

Therefore,

$$D = -\frac{1}{2} \times$$ (positive number) $$< 0$$

So, the determinant is negative.

Hence, the correct answer is Option B.