With usual notations, in triangle `A B C ,acos(B-C)+bcos(C-A)+c"cos"(A-B)` is equal to(a) `(a b c)/R^2` (b) `(a b c)/(4R^2)` (c) `(4a b c)/(R^2)` (d) `(a b c)/(2R^2)`

Explanation

To solve the problem, we need to evaluate the expression \( a \cos(B - C) + b \cos(C - A) + c \cos(A - B) \) in triangle \( ABC \) using the relationships between the sides and angles of the triangle.

### Step-by-Step Solution:

1. **Use the Law of Sines**:
We know from the Law of Sines that:
\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R
...