The number of distinct real values of `lamda`, for which the vectors `-lamda^(2)hati+hatj+hatk, hati-lamda^(2)hatj+hatk` and `hati+hatj-lamda^(2)hatk` are coplanar, is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To find the number of distinct real values of \( \lambda \) for which the vectors \( \mathbf{A} = -\lambda^2 \hat{i} + \hat{j} + \hat{k} \), \( \mathbf{B} = \hat{i} - \lambda^2 \hat{j} + \hat{k} \), and \( \mathbf{C} = \hat{i} + \hat{j} - \lambda^2 \hat{k} \) are coplanar, we need to calculate the scalar triple product of these vectors and set it equal to zero.

### Step 1: Write the vectors in component form
The vectors can be expressed as:
\[ \mathbf{A} = (-\lambda^2, 1, 1), \quad \mathbf{B} = (1, -\lambda^2, 1), \quad \mathbf{C} = (1, 1, -\lambda^2) \]

### Step 2: Set up the scalar triple product
The scalar triple product \( \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \) can be calculated using the determinant of a matrix formed by these vectors:
\[ \text{Det} = \begin{vmatrix} -\lambda^2 & 1 & 1 \\ 1 & -\lambda^2 & 1 \\ 1 & 1 & -\lambda^2 \end{vmatrix} \]

### Step 3: Calculate the determinant
We can expand this determinant:
\[ \text{Det} = -\lambda^2 \begin{vmatrix} -\lambda^2 & 1 \\ 1 & -\lambda^2 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & -\lambda^2 \end{vmatrix} + 1 \begin{vmatrix} 1 & -\lambda^2 \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -\lambda^2 & 1 \\ 1 & -\lambda^2 \end{vmatrix} = \lambda^4 - 1 \)
2. \( \begin{vmatrix} 1 & 1 \\ 1 & -\lambda^2 \end{vmatrix} = -\lambda^2 - 1 \)
3. \( \begin{vmatrix} 1 & -\lambda^2 \\ 1 & 1 \end{vmatrix} = 1 + \lambda^2 \)

Substituting back into the determinant:
\[ \text{Det} = -\lambda^2 (\lambda^4 - 1) + (\lambda^2 + 1) + (1 + \lambda^2) \]
\[ = -\lambda^6 + \lambda^2 + 1 + 1 + \lambda^2 \]
\[ = -\lambda^6 + 3\lambda^2 + 2 \]

### Step 4: Set the determinant to zero
To find the values of \( \lambda \) for which the vectors are coplanar, we set the determinant to zero:
\[ -\lambda^6 + 3\lambda^2 + 2 = 0 \]
Multiplying through by -1 gives:
\[ \lambda^6 - 3\lambda^2 - 2 = 0 \]

### Step 5: Substitute \( x = \lambda^2 \)
Let \( x = \lambda^2 \). The equation becomes:
\[ x^3 - 3x - 2 = 0 \]

### Step 6: Solve the cubic equation
We can use the Rational Root Theorem to test possible rational roots. Testing \( x = 2 \):
\[ 2^3 - 3(2) - 2 = 8 - 6 - 2 = 0 \]
So, \( x = 2 \) is a root. We can factor the cubic as:
\[ (x - 2)(x^2 + 2x + 1) = 0 \]
The quadratic factors as:
\[ (x - 2)(x + 1)^2 = 0 \]

### Step 7: Find the values of \( x \)
The solutions are:
1. \( x - 2 = 0 \) gives \( x = 2 \) (which means \( \lambda^2 = 2 \) leading to \( \lambda = \pm \sqrt{2} \))
2. \( (x + 1)^2 = 0 \) gives \( x = -1 \) (which does not yield real \( \lambda \))

### Step 8: Count distinct real values of \( \lambda \)
From \( x = 2 \), we have two distinct real values of \( \lambda \):
1. \( \lambda = \sqrt{2} \)
2. \( \lambda = -\sqrt{2} \)

Thus, the number of distinct real values of \( \lambda \) is **2**.

### Final Answer
The number of distinct real values of \( \lambda \) is \( \boxed{2} \).