If a, b, c are sides of the triangle ABC and `|(1,a, b),(1,c,a),(1,b,c)|=0`, then the value of `cos 2A+cos 2B+cos 2C` is equal to

Options

A. `-(3)/(2)`
B. `(3)/(2)`
C. `(3sqrt3)/(2)`
D. `-1`

Answer

To solve the problem, we need to find the value of \( \cos 2A + \cos 2B + \cos 2C \) given the condition \( |(1,a,b),(1,c,a),(1,b,c)|=0 \).

### Step-by-Step Solution:

1. **Understand the Determinant Condition**:
The condition \( |(1,a,b),(1,c,a),(1,b,c)|=0 \) implies that the rows of the matrix are linearly dependent. This means that the points represented by these rows are collinear.

2. **Expand the Determinant**:
We can expand the determinant:
\[ |(1,a,b),(1,c,a),(1,b,c)| = 1 \cdot (a \cdot c - b \cdot a) - 1 \cdot (b \cdot c - a \cdot b) + 1 \cdot (b \cdot a - c \cdot a) = 0 \]
Simplifying this, we get:
\[ ac - ab - bc + ab + ba - ca = 0 \implies a^2 + b^2 + c^2 - ab - ac - bc = 0 \]

3. **Rearranging the Equation**:
Rearranging gives us:
\[ a^2 + b^2 + c^2 = ab + ac + bc \]

4. **Recognizing the Triangle Condition**:
The equation \( a^2 + b^2 + c^2 = ab + ac + bc \) holds true if and only if \( a = b = c \). This indicates that triangle ABC is equilateral.

5. **Finding the Angles**:
In an equilateral triangle, all angles are equal:
\[ A = B = C = 60^\circ \]

6. **Calculating \( \cos 2A + \cos 2B + \cos 2C \)**:
We need to calculate:
\[ \cos 2A + \cos 2B + \cos 2C = \cos 120^\circ + \cos 120^\circ + \cos 120^\circ \]
Since \( \cos 120^\circ = -\frac{1}{2} \):
\[ \cos 2A + \cos 2B + \cos 2C = -\frac{1}{2} - \frac{1}{2} - \frac{1}{2} = -\frac{3}{2} \]

### Final Answer:
Thus, the value of \( \cos 2A + \cos 2B + \cos 2C \) is \( \boxed{-\frac{3}{2}} \).