If `a ne b ne c` are all positive, then the value of the determinant `|{:(a,b,c),(b,c,a),(c,a,b):}|`is

Options

A. non-negative
B. non-positive
C. negative
D. positive

Explanation

To find the value of the determinant

\[ D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]

we will follow these steps:

### Step 1: Apply Column Operations
We will perform a column operation by adding all three columns together. Specifically, we will replace the first column with the sum of the first, second, and third columns.

\[ C_1 \rightarrow C_1 + C_2 + C_3 \]

This gives us:

\[ D = \begin{vmatrix} a + b + c & b & c \\ a + b + c & c & a \\ a + b + c & a & b \end{vmatrix} \]

### Step 2: Factor Out the Common Term
Now, we can factor out \( (a + b + c) \) from the first column:

\[ D = (a + b + c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} \]

### Step 3: Calculate the 2x2 Determinant
Now we need to calculate the determinant of the 3x3 matrix:

\[ \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we can expand it:

\[ = 1 \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} 1 & a \\ 1 & b \end{vmatrix} + c \begin{vmatrix} 1 & c \\ 1 & a \end{vmatrix} \]

Calculating these 2x2 determinants:

1. \(\begin{vmatrix} c & a \\ a & b \end{vmatrix} = cb - a^2\)
2. \(\begin{vmatrix} 1 & a \\ 1 & b \end{vmatrix} = b - a\)
3. \(\begin{vmatrix} 1 & c \\ 1 & a \end{vmatrix} = a - c\)

Substituting these back, we get:

\[ D = (a + b + c) \left( cb - a^2 - b(b - a) + c(a - c) \right) \]

### Step 4: Simplify the Expression
Now, simplifying the expression inside the parentheses:

\[ = (a + b + c) \left( cb - a^2 - b^2 + ab + ac - c^2 \right) \]

### Step 5: Factor the Result
Notice that the expression \( cb - a^2 - b^2 + ab + ac - c^2 \) can be rewritten as:

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

### Step 6: Final Expression
Thus, we have:

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

Since \( a, b, c \) are all positive and distinct, \( (a-b)^2 + (b-c)^2 + (c-a)^2 > 0 \). Therefore, the determinant \( D \) is negative.

### Conclusion
Thus, the value of the determinant is negative.

\[ \text{Final Answer: } D < 0 \]