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 analyze the given condition and find the value of \( \cos 2A + \cos 2B + \cos 2C \).

### Step 1: Analyze the determinant condition
We are given that:

\[ \left| \begin{array}{ccc} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{array} \right| = 0 \]

This determinant being zero implies that the rows of the matrix are linearly dependent.

### Step 2: Perform row operations
We can perform row operations to simplify the determinant. Let's subtract the first row from the second and third rows:

\[ \left| \begin{array}{ccc} 1 & a & b \\ 0 & c-a & a-b \\ 0 & b-a & c-b \end{array} \right| \]

### Step 3: Expand the determinant
Now we can expand the determinant:

\[ = 1 \cdot \left| \begin{array}{cc} c-a & a-b \\ b-a & c-b \end{array} \right| \]

Calculating this 2x2 determinant gives us:

\[ = (c-a)(c-b) - (a-b)(b-a) = (c-a)(c-b) - (a-b)^2 \]

### Step 4: Set the determinant to zero
Setting the determinant equal to zero gives us:

\[ (c-a)(c-b) - (a-b)^2 = 0 \]

### Step 5: Rearranging the equation
This can be rearranged to:

\[ (c-a)(c-b) = (a-b)^2 \]

### Step 6: Analyze the implications
For this equation to hold true, since all terms are positive (as they represent sides of a triangle), we can conclude that:

\[ a = b = c \]

This means triangle ABC is equilateral.

### Step 7: Find angles A, B, and C
In an equilateral triangle, all angles are equal:

\[ A = B = C = \frac{\pi}{3} \]

### Step 8: Calculate \( \cos 2A + \cos 2B + \cos 2C \)
Now, we need to find:

\[ \cos 2A + \cos 2B + \cos 2C \]

Since \( A = B = C = \frac{\pi}{3} \):

\[ \cos 2A = \cos 2B = \cos 2C = \cos \left(2 \cdot \frac{\pi}{3}\right) = \cos \left(\frac{2\pi}{3}\right) = -\frac{1}{2} \]

Thus,

\[ \cos 2A + \cos 2B + \cos 2C = 3 \cdot \left(-\frac{1}{2}\right) = -\frac{3}{2} \]

### Final Answer
The value of \( \cos 2A + \cos 2B + \cos 2C \) is:

\[ \boxed{-\frac{3}{2}} \]