Suppose a,b,c are sides of a scalene triangle. Let
`Delta=|(a,b,c),(b,c,a),(c,a,b)|`
Then

Options

A. `Deltale0`
B. `Deltalt0`
C. `Deltagt0`
D. `Deltage0`

Answer

To solve the problem, we need to evaluate the determinant \( \Delta = |(a,b,c),(b,c,a),(c,a,b)| \) where \( a, b, c \) are the sides of a scalene triangle.

### Step-by-step Solution:

1. **Write the Determinant**:
We start with the determinant:
\[ \Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]

2. **Expand the Determinant**:
We can expand this determinant using the rule of Sarrus or cofactor expansion. For this determinant, we can use the first row:
\[ \Delta = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

3. **Calculate the 2x2 Determinants**:
Now we calculate each of the 2x2 determinants:
- For the first determinant:
\[ \begin{vmatrix} c & a \\ a & b \end{vmatrix} = cb - a^2 \]
- For the second determinant:
\[ \begin{vmatrix} b & a \\ c & b \end{vmatrix} = bb - ac = b^2 - ac \]
- For the third determinant:
\[ \begin{vmatrix} b & c \\ c & a \end{vmatrix} = ba - c^2 \]

4. **Substituting Back**:
Substitute these back into the expression for \( \Delta \):
\[ \Delta = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) \]
Simplifying this gives:
\[ \Delta = acb - a^3 - b^3 + abc + abc - c^3 \]
Thus,
\[ \Delta = 3abc - (a^3 + b^3 + c^3) \]

5. **Using the Identity**:
We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Therefore,
\[ \Delta = - (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

6. **Analyzing the Result**:
Since \( a, b, c \) are the sides of a scalene triangle, we know:
- \( a + b + c > 0 \)
- \( a^2 + b^2 + c^2 - ab - ac - bc \) is non-negative (it is zero if \( a = b = c \), which is not the case here).

Thus, \( \Delta < 0 \).

### Conclusion:
Therefore, we conclude that:
\[ \Delta < 0 \]