if `a+b+c=0` then `Delta=|{:(a-x,c,b),(c,b-x,a),(b,a,c-x):}|=0` is

Options

A. `x=0`
B. `x=sqrt(a^(2)+b^(2)+c^(2))`
C. `x=+sqrt((3)/(2)(a^(2)+b^(2)+c^(2)))`
D. `x=-sqrt((3)/(2)(a^(2)+b^(2)+c^(2)))`

Explanation

To solve the problem, we need to evaluate the determinant given the condition \( a + b + c = 0 \) and find the values of \( x \) that make the determinant equal to zero.

### Step-by-Step Solution:

1. **Write the Determinant**:
\[ \Delta = \begin{vmatrix} a - x & c & b \\ c & b - x & a \\ b & a & c - x \end{vmatrix} \]

2. **Add the Columns**:
We will add the three columns together:
\[ C_1 \to C_1 + C_2 + C_3 \]
This gives us:
\[ \Delta = \begin{vmatrix} (a + b + c - x) & c & b \\ (a + b + c) & (b - x) & a \\ (a + b + c) & a & (c - x) \end{vmatrix} \]

3. **Substituting \( a + b + c = 0 \)**:
Since \( a + b + c = 0 \), we can substitute this into the determinant:
\[ \Delta = \begin{vmatrix} -x & c & b \\ 0 & b - x & a \\ 0 & a & -x \end{vmatrix} \]

4. **Factor Out**:
The first column can be factored out:
\[ \Delta = -x \begin{vmatrix} 1 & c & b \\ 0 & b - x & a \\ 0 & a & -x \end{vmatrix} \]

5. **Calculate the Determinant**:
The determinant simplifies to:
\[ \Delta = -x \left[ (b - x)(-x) - a \cdot c \right] \]
Expanding this gives:
\[ \Delta = -x \left[ -bx + x^2 - ac \right] \]

6. **Set the Determinant to Zero**:
We set the determinant equal to zero:
\[ -x(-bx + x^2 - ac) = 0 \]
This gives us two cases:
- \( x = 0 \)
- \( -bx + x^2 - ac = 0 \)

7. **Solve the Quadratic Equation**:
Rearranging the second equation:
\[ x^2 - bx - ac = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2} \).

8. **Final Values of \( x \)**:
Thus, the values of \( x \) are:
\[ x = 0 \quad \text{and} \quad x = \frac{b \pm \sqrt{b^2 + 4ac}}{2} \]