If `a+b+c=0`, then one root of `|(a-n,c,b),(c,b-n,a),(b,a,c-n)|=0` is

Options

A. `n=1`
B. `n=2`
C. `n=a^(2)+b^(2)+c^(2)`
D. `n=0`

Answer

To solve the problem, we need to find one root of the determinant equation given that \( a + b + c = 0 \). The determinant we are dealing with is:

\[ \begin{vmatrix} a - n & c & b \\ c & b - n & a \\ b & a & c - n \end{vmatrix} = 0 \]

### Step-by-Step Solution:

**Step 1: Substitute \( a + b + c = 0 \)**

Since \( a + b + c = 0 \), we can express \( c \) as \( c = -a - b \).

**Step 2: Write the determinant**

The determinant can be rewritten as:

\[ D = \begin{vmatrix} a - n & -a - b & b \\ -a - b & b - n & a \\ b & a & -a - b - n \end{vmatrix} \]

**Step 3: Perform row operations**

We can simplify the determinant by performing row operations. Let's add the first row to the second and third rows:

\[ D = \begin{vmatrix} a - n & -a - b & b \\ 0 & (b - n) + (-a - b) & a + b \\ 0 & a + b & -a - b - n \end{vmatrix} \]

This simplifies to:

\[ D = \begin{vmatrix} a - n & -a - b & b \\ 0 & -a - n & a + b \\ 0 & a + b & -a - b - n \end{vmatrix} \]

**Step 4: Factor out common terms**

Now, we can factor out \( (a + b) \) from the second and third rows:

\[ D = (a + b) \begin{vmatrix} a - n & -a - b & b \\ 0 & -a - n & 1 \\ 0 & 1 & -1 - \frac{n}{a + b} \end{vmatrix} \]

**Step 5: Expand the determinant**

Now we can expand the determinant. The determinant simplifies to:

\[ D = (a + b) \cdot (a - n) \cdot \left( (-a - n)(-1 - \frac{n}{a + b}) - 1 \right) \]

**Step 6: Set the determinant to zero**

For the determinant to be zero, we have two cases:

1. \( a + b = 0 \)
2. \( a - n = 0 \)

From the first case \( a + b = 0 \), we can substitute \( c = -a - b \) which leads to \( c = 0 \).

From the second case \( a - n = 0 \), we have \( n = a \).

**Step 7: Solve for \( n \)**

Since \( a + b + c = 0 \), we can conclude that \( n = 0 \) is a root.

### Conclusion

Thus, one root of the equation \( |(a-n,c,b),(c,b-n,a),(b,a,c-n)| = 0 \) is:

\[ \boxed{0} \]