If `|{:(1,cos alpha, cos beta),(cos alpha, 1 , cos gamma ),(cos beta, cos gamma , 1):}|=|{:(0,cos alpha, cos beta),(cos alpha , 0 , cos gamma),(cos beta, cos gamma, 0):}|` then the value of `cos^2 alpha + cos^2 beta + cos^2 gamma` is :

Options

A. 1
B. `1/2`
C. `3/8`
D. `9/4`

Answer

To solve the given problem, we need to evaluate the determinants and set them equal to each other. Let's denote the determinants as \( D_1 \) and \( D_2 \).

### Step 1: Write the Determinants
The first determinant \( D_1 \) is given by:

\[ D_1 = \begin{vmatrix} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{vmatrix} \]

The second determinant \( D_2 \) is given by:

\[ D_2 = \begin{vmatrix} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{vmatrix} \]

### Step 2: Calculate \( D_1 \)
To calculate \( D_1 \), we can use the formula for the determinant of a 3x3 matrix:

\[ D_1 = 1 \cdot \begin{vmatrix} 1 & \cos \gamma \\ \cos \gamma & 1 \end{vmatrix} - \cos \alpha \cdot \begin{vmatrix} \cos \alpha & \cos \gamma \\ \cos \beta & 1 \end{vmatrix} + \cos \beta \cdot \begin{vmatrix} \cos \alpha & 1 \\ \cos \beta & \cos \gamma \end{vmatrix} \]

Calculating the first term:

\[ \begin{vmatrix} 1 & \cos \gamma \\ \cos \gamma & 1 \end{vmatrix} = 1 - \cos^2 \gamma = \sin^2 \gamma \]

Calculating the second term:

\[ \begin{vmatrix} \cos \alpha & \cos \gamma \\ \cos \beta & 1 \end{vmatrix} = \cos \alpha \cdot 1 - \cos \beta \cdot \cos \gamma = \cos \alpha - \cos \beta \cos \gamma \]

Calculating the third term:

\[ \begin{vmatrix} \cos \alpha & 1 \\ \cos \beta & \cos \gamma \end{vmatrix} = \cos \alpha \cdot \cos \gamma - \cos \beta \cdot 1 = \cos \alpha \cos \gamma - \cos \beta \]

Putting it all together:

\[ D_1 = 1 \cdot \sin^2 \gamma - \cos \alpha (\cos \alpha - \cos \beta \cos \gamma) + \cos \beta (\cos \alpha \cos \gamma - \cos \beta) \]

Expanding this gives:

\[ D_1 = \sin^2 \gamma - \cos^2 \alpha + \cos \alpha \cos \beta \cos \gamma + \cos \beta \cos \alpha \cos \gamma - \cos^2 \beta \]

So,

\[ D_1 = \sin^2 \gamma - \cos^2 \alpha - \cos^2 \beta + 2 \cos \alpha \cos \beta \cos \gamma \]

### Step 3: Calculate \( D_2 \)
For \( D_2 \):

\[ D_2 = -\begin{vmatrix} \cos \alpha & \cos \beta \\ \cos \beta & \cos \gamma \end{vmatrix} = -(\cos \alpha \cos \gamma - \cos^2 \beta) \]

So,

\[ D_2 = -\cos \alpha \cos \gamma + \cos^2 \beta \]

### Step 4: Set \( D_1 = D_2 \)
Now we set \( D_1 = D_2 \):

\[ \sin^2 \gamma - \cos^2 \alpha - \cos^2 \beta + 2 \cos \alpha \cos \beta \cos \gamma = -\cos \alpha \cos \gamma + \cos^2 \beta \]

### Step 5: Simplify the Equation
Rearranging gives:

\[ \sin^2 \gamma - \cos^2 \alpha + \cos \alpha \cos \gamma = 0 \]

Using the identity \( \sin^2 \gamma = 1 - \cos^2 \gamma \):

\[ 1 - \cos^2 \gamma - \cos^2 \alpha + \cos \alpha \cos \gamma = 0 \]

### Step 6: Solve for \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma \)
From the equation, we can derive:

\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

### Final Answer
Thus, the value of \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma \) is:

\[ \boxed{1} \]