Let `A= [[a,b,c],[b,c,a],[c,a,b]]` is an orthogonal matrix and `abc = lambda (lt0).`
The equation whose roots are `a, b, c, ` is

Options

A. `x^(3)- 2x^(2) + lambda = 0`
B. `x^(3) -lambda x^(2) + lambda x + lambda = 0 `
C. `x^(3) - 2 x^(2) + 2 lambda x + lambda = 0 `
D. `x^(3) pm x^(2) - lambda = 0 `

Explanation

To solve the problem, we need to find the equation whose roots are \(a\), \(b\), and \(c\) given that the matrix

\[ A = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \]

is an orthogonal matrix and \(abc = \lambda\) (where \(\lambda < 0\)).

### Step 1: Use the property of orthogonal matrices
An orthogonal matrix satisfies the condition \(A A^T = I\), where \(I\) is the identity matrix.

### Step 2: Compute \(A^T\)
The transpose of matrix \(A\) is:

\[ A^T = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \]

Since \(A\) is symmetric, we have \(A^T = A\).

### Step 3: Set up the equation
We need to compute \(A A^T\):

\[ A A^T = A^2 = \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \begin{bmatrix} a & b & c \\ b & c & a \\ c & a & b \end{bmatrix} \]

### Step 4: Perform the multiplication
Calculating the elements of \(A^2\):

1. First row, first column:
\[ a^2 + b^2 + c^2 \]
2. First row, second column:
\[ ab + bc + ca \]
3. First row, third column:
\[ ac + ba + cb \]

Repeating this for the other rows, we find:

\[ A^2 = \begin{bmatrix} a^2 + b^2 + c^2 & ab + bc + ca & ab + bc + ca \\ ab + bc + ca & b^2 + c^2 + a^2 & ac + ba + cb \\ ac + ba + cb & ac + ba + cb & c^2 + a^2 + b^2 \end{bmatrix} \]

### Step 5: Set the equation equal to the identity matrix
Since \(A\) is orthogonal, we have:

\[ A^2 = I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

From this, we can equate the corresponding elements:

1. Diagonal elements:
\[ a^2 + b^2 + c^2 = 1 \]
2. Off-diagonal elements:
\[ ab + bc + ca = 0 \]

### Step 6: Form a polynomial equation
We need to find the polynomial whose roots are \(a\), \(b\), and \(c\). The polynomial can be expressed as:

\[ x^3 - (a+b+c)x^2 + (ab + ac + bc)x - abc = 0 \]

### Step 7: Substitute known values
From the previous steps, we know:

- \(ab + ac + bc = 0\)
- Let \(s = a + b + c\)

Thus, the polynomial simplifies to:

\[ x^3 - sx^2 - \lambda = 0 \]

### Step 8: Determine the value of \(s\)
Since \(a^2 + b^2 + c^2 = 1\), we can use the identity:

\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]

Substituting the known values:

\[ s^2 = 1 + 2(0) \Rightarrow s^2 = 1 \Rightarrow s = \pm 1 \]

### Step 9: Final polynomial
Thus, the equation whose roots are \(a\), \(b\), and \(c\) is:

\[ x^3 \mp x^2 - \lambda = 0 \]

### Conclusion
The final polynomial equation is:

\[ x^3 - sx^2 - \lambda = 0 \]