Let `veca=hati+hatj+hatk, vecb=hati-hatj+hat2k` and `vecc=xhati+(x-2)hatj-hatk`. If the vector `vecc` lies in the plane of `veca` and `vecb` then `x` equals

Options

A. 0
B. 1
C. `-4`
D. `-2`

Answer

To solve the problem, we need to find the value of \( x \) such that the vector \( \vec{c} \) lies in the plane formed by the vectors \( \vec{a} \) and \( \vec{b} \). This can be determined using the condition of coplanarity, which states that if three vectors are coplanar, their scalar triple product is zero.

### Step-by-Step Solution:

1. **Identify the Vectors**:
- Given:
\[ \vec{a} = \hat{i} + \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \]
\[ \vec{c} = x\hat{i} + (x-2)\hat{j} - \hat{k} \]

2. **Set Up the Scalar Triple Product**:
- The scalar triple product of vectors \( \vec{a}, \vec{b}, \vec{c} \) can be represented using the determinant:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0 \]

3. **Calculate the Determinant**:
- We will simplify the determinant step by step:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} \]
- Perform column operations:
- Column 1: \( C_1 \leftarrow C_1 - C_2 \)
- Column 3: \( C_3 \leftarrow C_3 - C_2 \)
\[ \begin{vmatrix} 0 & 1 & 2 \\ 2 & -1 & 3 \\ x+2 & x-2 & 1-x \end{vmatrix} \]

4. **Expand the Determinant**:
- Now we will expand the determinant:
\[ = 0 \cdot \begin{vmatrix} -1 & 3 \\ x-2 & 1-x \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 3 \\ x+2 & 1-x \end{vmatrix} + 2 \cdot \begin{vmatrix} 2 & -1 \\ x+2 & x-2 \end{vmatrix} \]
- Calculate the 2x2 determinants:
\[ = -1 \left( 2(1-x) - 3(x+2) \right) + 2 \left( 2(x-2) + 1(x+2) \right) \]
- Simplifying gives:
\[ = -1(2 - 2x - 3x - 6) + 2(2x - 4 + x + 2) \]
\[ = -1(-5x - 4) + 2(3x - 2) \]
\[ = 5x + 4 + 6x - 4 = 11x \]

5. **Set the Determinant to Zero**:
- For coplanarity, set the determinant equal to zero:
\[ 11x = 0 \]
- Thus, solving for \( x \):
\[ x = 0 \]

### Final Answer:
The value of \( x \) is \( \boxed{0} \).