Number of integral value(s) of `lambda` for which vectors `x^(2)hati-hatj+xhatk, (lambda-1)hati-2lambdahatj-hatk` and `hati-hatj+hatk`, in the order form right-handed system `AA x` `in`R, is
(a) 0 (b) 2 (c) 4 (d) 6

Options

A. 0
B. 2
C. 4
D. 6

Explanation

To solve the problem of finding the number of integral values of \( \lambda \) for which the vectors \( \mathbf{a} = x^2 \hat{i} - \hat{j} + x \hat{k} \), \( \mathbf{b} = (\lambda - 1) \hat{i} - 2\lambda \hat{j} - \hat{k} \), and \( \mathbf{c} = \hat{i} - \hat{j} + \hat{k} \) form a right-handed system, we need to calculate the scalar triple product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \) and ensure that it is greater than zero.

### Step 1: Set up the vectors
We have:
- \( \mathbf{a} = x^2 \hat{i} - \hat{j} + x \hat{k} \)
- \( \mathbf{b} = (\lambda - 1) \hat{i} - 2\lambda \hat{j} - \hat{k} \)
- \( \mathbf{c} = \hat{i} - \hat{j} + \hat{k} \)

### Step 2: Calculate the cross product \( \mathbf{b} \times \mathbf{c} \)
To find \( \mathbf{b} \times \mathbf{c} \), we can use the determinant of a matrix formed by the unit vectors and the components of \( \mathbf{b} \) and \( \mathbf{c} \):

\[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda - 1 & -2\lambda & -1 \\ 1 & -1 & 1 \end{vmatrix} \]

Calculating this determinant:

\[ \mathbf{b} \times \mathbf{c} = \hat{i} \begin{vmatrix} -2\lambda & -1 \\ -1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} \lambda - 1 & -1 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} \lambda - 1 & -2\lambda \\ 1 & -1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} -2\lambda & -1 \\ -1 & 1 \end{vmatrix} = (-2\lambda)(1) - (-1)(-1) = -2\lambda - 1 \)

2. \( \begin{vmatrix} \lambda - 1 & -1 \\ 1 & 1 \end{vmatrix} = (\lambda - 1)(1) - (-1)(1) = \lambda - 1 + 1 = \lambda \)

3. \( \begin{vmatrix} \lambda - 1 & -2\lambda \\ 1 & -1 \end{vmatrix} = (\lambda - 1)(-1) - (-2\lambda)(1) = -\lambda + 1 + 2\lambda = \lambda + 1 \)

Combining these results, we have:

\[ \mathbf{b} \times \mathbf{c} = (-2\lambda - 1) \hat{i} - \lambda \hat{j} + (\lambda + 1) \hat{k} \]

### Step 3: Calculate the dot product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \)
Now we compute the dot product:

\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (x^2 \hat{i} - \hat{j} + x \hat{k}) \cdot [(-2\lambda - 1) \hat{i} - \lambda \hat{j} + (\lambda + 1) \hat{k}] \]

Calculating this gives:

\[ = x^2(-2\lambda - 1) + (-1)(-\lambda) + x(\lambda + 1) \]

Simplifying this:

\[ = -2\lambda x^2 - x^2 + \lambda + x(\lambda + 1) = -2\lambda x^2 - x^2 + \lambda + \lambda x + x \]

Combining like terms:

\[ = -2\lambda x^2 + (\lambda + 1)x + \lambda \]

### Step 4: Set the scalar triple product greater than zero
We need:

\[ -2\lambda x^2 + (\lambda + 1)x + \lambda > 0 \]

### Step 5: Analyze the quadratic inequality
This is a quadratic in \( x \). For the quadratic to be positive, we need to consider the conditions on \( \lambda \):

1. The coefficient of \( x^2 \) must be negative: \( -2\lambda < 0 \Rightarrow \lambda > 0 \).
2. The discriminant must be positive for the quadratic to have real roots:

\[ D = (\lambda + 1)^2 - 4(-2\lambda)(\lambda) > 0 \]

Calculating the discriminant:

\[ D = (\lambda + 1)^2 + 8\lambda^2 = 9\lambda^2 + 2\lambda + 1 > 0 \]

This is always positive since it is a perfect square.

### Step 6: Combine conditions
From the conditions derived:

- \( \lambda > 0 \)

### Step 7: Find integral values
The integral values of \( \lambda \) satisfying \( \lambda > 0 \) are \( 1, 2, 3, \ldots \). However, we need to check if there are any upper bounds based on the original conditions.

Since the quadratic must be positive for all \( x \), we conclude that there are no upper bounds on \( \lambda \).

### Conclusion
Thus, the number of integral values of \( \lambda \) for which the vectors form a right-handed system is infinite.

### Final Answer
(a) 0