Let `vec(a) = hat(i) + hat(j) + hat(k), vec(b) = hat(i) - hat(j) + 2hat(k)` and `vec(c) = xhat(i) + (x-2)hat(j) - hat(k)`. If the vector `vec(c)` lies in the plane of `vec(a)` & `vec(b)`, then x equals

Options

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

Answer

To solve the problem, we need to determine the value of \( x \) such that the vector \( \vec{c} \) lies in the plane formed by the vectors \( \vec{a} \) and \( \vec{b} \).

### Step-by-step Solution:

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

2. **Condition for Coplanarity:**
- Vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are coplanar if the determinant of the matrix formed by their components is zero:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0 \]

3. **Calculate the Determinant:**
- We can calculate the determinant using the formula:
\[ \text{Det} = 1 \cdot \begin{vmatrix} -1 & 2 \\ x-2 & -1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 2 \\ x & -1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -1 \\ x & x-2 \end{vmatrix} \]

- Calculate each of the 2x2 determinants:
- First determinant:
\[ \begin{vmatrix} -1 & 2 \\ x-2 & -1 \end{vmatrix} = (-1)(-1) - (2)(x-2) = 1 - 2x + 4 = 5 - 2x \]
- Second determinant:
\[ \begin{vmatrix} 1 & 2 \\ x & -1 \end{vmatrix} = (1)(-1) - (2)(x) = -1 - 2x = -1 - 2x \]
- Third determinant:
\[ \begin{vmatrix} 1 & -1 \\ x & x-2 \end{vmatrix} = (1)(x-2) - (-1)(x) = x - 2 + x = 2x - 2 \]

- Combine these results:
\[ \text{Det} = 1(5 - 2x) - 1(-1 - 2x) + 1(2x - 2) \]
\[ = 5 - 2x + 1 + 2x + 2x - 2 \]
\[ = 5 + 1 - 2 + 2x - 2x = 4 \]

4. **Set the Determinant to Zero:**
- Set the determinant equal to zero:
\[ 4 = 0 \]
- This is incorrect; let's re-evaluate the determinant calculation.

5. **Re-evaluate the Determinant:**
- The correct determinant calculation should yield:
\[ 5 - 2x + 1 + 2x + 2x - 2 = 0 \]
\[ 4 + 2x = 0 \]
\[ 2x = -4 \]
\[ x = -2 \]

### Final Answer:
Thus, the value of \( x \) is:
\[ \boxed{-2} \]