if `f (x) = |{:(x,,a,,a),(a,,x,,a),(a,,a,,x):}|=0` then

Options

A. `f(x) =0" and " f(x) =0` has one common root
B. `f(x) =0" and " f(x) =0` has one common root
C. sum of roots of f(x) =0 is -3a
D. none of these

Explanation

To solve the problem, we need to evaluate the determinant given in the function \( f(x) = \begin{vmatrix} x & a & a \\ a & x & a \\ a & a & x \end{vmatrix} \) and find the values of \( x \) for which \( f(x) = 0 \).

### Step-by-Step Solution:

1. **Write the Determinant**:
\[ f(x) = \begin{vmatrix} x & a & a \\ a & x & a \\ a & a & x \end{vmatrix} \]

2. **Add the Columns**:
We can simplify the determinant by adding all the columns together. Let’s replace the first column \( C_1 \) with \( C_1 + C_2 + C_3 \):
\[ C_1 \rightarrow C_1 + C_2 + C_3 \]
This gives us:
\[ f(x) = \begin{vmatrix} x + 2a & a & a \\ a & x & a \\ a & a & x \end{vmatrix} \]

3. **Factor Out Common Terms**:
We can factor out \( (x + 2a) \) from the first column:
\[ f(x) = (x + 2a) \begin{vmatrix} 1 & a & a \\ a & x & a \\ a & a & x \end{vmatrix} \]

4. **Subtract Rows**:
To simplify the determinant further, we can subtract the first row from the second and third rows:
\[ R_2 \rightarrow R_2 - R_1 \quad \text{and} \quad R_3 \rightarrow R_3 - R_1 \]
This results in:
\[ f(x) = (x + 2a) \begin{vmatrix} 1 & a & a \\ 0 & x - a & 0 \\ 0 & 0 & x - a \end{vmatrix} \]

5. **Calculate the Determinant**:
The determinant can now be calculated:
\[ f(x) = (x + 2a) \cdot 1 \cdot (x - a)(x - a) = (x + 2a)(x - a)^2 \]

6. **Set the Function to Zero**:
To find the roots, we set \( f(x) = 0 \):
\[ (x + 2a)(x - a)^2 = 0 \]

7. **Find the Roots**:
The roots are found by solving each factor:
- From \( x + 2a = 0 \):
\[ x = -2a \]
- From \( (x - a)^2 = 0 \):
\[ x = a \quad (\text{with multiplicity 2}) \]

### Summary of Roots:
The roots of the equation are:
- \( x = a \) (with multiplicity 2)
- \( x = -2a \) (with multiplicity 1)

### Final Answer:
The roots are \( a, a, -2a \).