The number of distinct real values of `lambda`, for which the vectors `-lambda^(2)hat(i)+hat(j)+hat(k), hat(i)-lambda^(2)hat(j)+hat(k) and hat(i)+hat(j)-lambda^(2)hat(k)` are coplanar, is

Options

A. `0`
B. `1`
C. `pmsqrt(2)`
D. `3`

Answer

To determine the number of distinct real values of \(\lambda\) for which the vectors \(-\lambda^2 \hat{i} + \hat{j} + \hat{k}\), \(\hat{i} - \lambda^2 \hat{j} + \hat{k}\), and \(\hat{i} + \hat{j} - \lambda^2 \hat{k}\) are coplanar, we can follow these steps:

### Step 1: Define the Vectors
Let:
\[ \mathbf{a} = -\lambda^2 \hat{i} + \hat{j} + \hat{k} \]
\[ \mathbf{b} = \hat{i} - \lambda^2 \hat{j} + \hat{k} \]
\[ \mathbf{c} = \hat{i} + \hat{j} - \lambda^2 \hat{k} \]

### Step 2: Set Up the Determinant
The vectors are coplanar if the determinant of the matrix formed by these vectors is zero:
\[ \begin{vmatrix} -\lambda^2 & 1 & 1 \\ 1 & -\lambda^2 & 1 \\ 1 & 1 & -\lambda^2 \end{vmatrix} = 0 \]

### Step 3: Calculate the Determinant
We can calculate the determinant using cofactor expansion or row operations. Let's perform row operations to simplify the determinant:
1. Add the first row to the second and third rows:
\[ \begin{vmatrix} -\lambda^2 & 1 & 1 \\ 0 & -\lambda^2 + 1 & 2 \\ 0 & 2 & -\lambda^2 + 1 \end{vmatrix} \]

### Step 4: Further Simplify the Determinant
Now, we can factor out \(-\lambda^2\):
\[ -\lambda^2 \begin{vmatrix} 1 & 1 & 1 \\ 0 & -\lambda^2 + 1 & 2 \\ 0 & 2 & -\lambda^2 + 1 \end{vmatrix} \]

### Step 5: Calculate the Remaining 2x2 Determinant
Now, we calculate the determinant of the remaining 2x2 matrix:
\[ \begin{vmatrix} -\lambda^2 + 1 & 2 \\ 2 & -\lambda^2 + 1 \end{vmatrix} = (-\lambda^2 + 1)(-\lambda^2 + 1) - (2)(2) = (\lambda^2 - 1)^2 - 4 \]
This simplifies to:
\[ (\lambda^2 - 1)^2 - 4 = 0 \]

### Step 6: Solve the Equation
Now we have:
\[ (\lambda^2 - 1)^2 - 4 = 0 \]
This can be factored as:
\[ (\lambda^2 - 1 - 2)(\lambda^2 - 1 + 2) = 0 \]
which gives:
\[ (\lambda^2 - 3)(\lambda^2 + 1) = 0 \]

### Step 7: Find the Values of \(\lambda\)
From \((\lambda^2 - 3) = 0\), we get:
\[ \lambda^2 = 3 \implies \lambda = \pm \sqrt{3} \]
From \((\lambda^2 + 1) = 0\), we get:
\[ \lambda^2 = -1 \implies \text{no real solutions} \]

### Conclusion
Thus, the distinct real values of \(\lambda\) are \(\sqrt{3}\) and \(-\sqrt{3}\). Therefore, the number of distinct real values of \(\lambda\) is:
\[ \boxed{2} \]