The value of the determinant
`|(1,cos(B-A),cos(C-A)),(cos(A-B),1,cos(C-B)),(cos(A-C),cos(B-C),1)|` is

Options

A. `4cos A cos B cos C`
B. `2cos A cos B cos C`
C. `4sin A sin B sinC`
D. zero

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} 1 & \cos(B-A) & \cos(C-A) \\ \cos(A-B) & 1 & \cos(C-B) \\ \cos(A-C) & \cos(B-C) & 1 \end{vmatrix} \]

we will use properties of determinants and trigonometric identities.

### Step 1: Rewrite the Cosine Terms
Using the cosine subtraction formula, we can express the cosine terms in the determinant:

\[ \cos(B-A) = \cos B \cos A + \sin B \sin A \]
\[ \cos(A-B) = \cos A \cos B + \sin A \sin B \]
\[ \cos(C-A) = \cos C \cos A + \sin C \sin A \]
\[ \cos(A-C) = \cos A \cos C + \sin A \sin C \]
\[ \cos(C-B) = \cos C \cos B + \sin C \sin B \]
\[ \cos(B-C) = \cos B \cos C + \sin B \sin C \]

### Step 2: Substitute into the Determinant
Substituting these identities into the determinant gives:

\[ D = \begin{vmatrix} 1 & \cos B \cos A + \sin B \sin A & \cos C \cos A + \sin C \sin A \\ \cos A \cos B + \sin A \sin B & 1 & \cos C \cos B + \sin C \sin B \\ \cos A \cos C + \sin A \sin C & \cos B \cos C + \sin B \sin C & 1 \end{vmatrix} \]

### Step 3: Simplify the Determinant
Notice that the determinant has a symmetric structure. We can also observe that if we replace one row with the sum of the rows, the value of the determinant remains unchanged.

### Step 4: Apply the Determinant Properties
Using the property of determinants where if two rows (or columns) are identical, the determinant is zero, we can see that if we add the first row to the second and third rows, we will create identical rows, leading to:

\[ D = 0 \]

### Conclusion
Thus, the value of the determinant is:

\[ \boxed{0} \]