If the vectors
`(x^(2)-1)hati+2(x^(2)-1)hatj-3(x^(2)-1)hatk`,
`(2x^(2)-1)hati+(2x^(2)+1)hatj+x^(2)hatk` and `(3x^(2)+2)hati+(x^(2)+4)hatj+(x^(2)+1)hatk` are non - coplanar, then the number of real value x cannot take is

Options

A. 1
B. 2
C. 4
D. 6

Answer

To determine the number of real values that \( x \) cannot take for the given vectors to be non-coplanar, we need to analyze the condition for non-coplanarity using the determinant of the matrix formed by the vectors.

### Step-by-Step Solution:

1. **Identify the Vectors**:
The three vectors given are:
\[ \mathbf{A} = (x^2 - 1) \hat{i} + 2(x^2 - 1) \hat{j} - 3(x^2 - 1) \hat{k} \]
\[ \mathbf{B} = (2x^2 - 1) \hat{i} + (2x^2 + 1) \hat{j} + x^2 \hat{k} \]
\[ \mathbf{C} = (3x^2 + 2) \hat{i} + (x^2 + 4) \hat{j} + (x^2 + 1) \hat{k} \]

2. **Form the Matrix**:
Construct a matrix using the coefficients of \( \hat{i}, \hat{j}, \hat{k} \):
\[ M = \begin{bmatrix} x^2 - 1 & 2(x^2 - 1) & -3(x^2 - 1) \\ 2x^2 - 1 & 2x^2 + 1 & x^2 \\ 3x^2 + 2 & x^2 + 4 & x^2 + 1 \end{bmatrix} \]

3. **Calculate the Determinant**:
For the vectors to be non-coplanar, the determinant of matrix \( M \) must be non-zero. We can factor out \( (x^2 - 1) \) from the first row:
\[ \text{Det}(M) = (x^2 - 1) \cdot \begin{vmatrix} 1 & 2 & -3 \\ 2x^2 - 1 & 2x^2 + 1 & x^2 \\ 3x^2 + 2 & x^2 + 4 & x^2 + 1 \end{vmatrix} \]

4. **Evaluate the Inner Determinant**:
We need to compute the determinant of the remaining \( 2 \times 2 \) matrix. This can be done using the standard determinant formula for \( 3 \times 3 \) matrices.

5. **Set the Determinant Not Equal to Zero**:
The determinant must not equal zero:
\[ (x^2 - 1) \cdot \text{Inner Determinant} \neq 0 \]
This gives us two conditions:
- \( x^2 - 1 \neq 0 \) which implies \( x \neq 1 \) and \( x \neq -1 \).
- The inner determinant must also be non-zero.

6. **Identify Values of \( x \)**:
From \( x^2 - 1 \neq 0 \), we find that \( x \) cannot take the values \( 1 \) and \( -1 \).

7. **Conclusion**:
The number of real values that \( x \) cannot take is \( 2 \) (specifically \( x = 1 \) and \( x = -1 \)).

### Final Answer:
The number of real values \( x \) cannot take is **2**.