Let `veca=hati+2hatj+3hati, vecb=hati-hatj+2hatk and vecc=(x-2)hati-(x-3)hatj-hatk`. If `vecc` lies in the plane of `veca and vecb`, then `(1)/(x)` is equal to

Answer

To solve the problem, we need to determine the value of \( \frac{1}{x} \) given that the vector \( \vec{c} \) lies in the plane of vectors \( \vec{a} \) and \( \vec{b} \). This means that the vectors \( \vec{a}, \vec{b}, \) and \( \vec{c} \) are coplanar, which can be established using the determinant of a matrix formed by these vectors.

### Step-by-Step Solution:

1. **Identify the vectors**:
- \( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \)
- \( \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \)
- \( \vec{c} = (x-2)\hat{i} - (x-3)\hat{j} - \hat{k} \)

2. **Write the coefficients of the vectors**:
- Coefficients of \( \vec{a} \): \( (1, 2, 3) \)
- Coefficients of \( \vec{b} \): \( (1, -1, 2) \)
- Coefficients of \( \vec{c} \): \( (x-2, -(x-3), -1) \)

3. **Set up the determinant**:
The condition for coplanarity is given by the determinant:
\[ \begin{vmatrix} 1 & 2 & 3 \\ 1 & -1 & 2 \\ x-2 & -(x-3) & -1 \end{vmatrix} = 0 \]

4. **Calculate the determinant**:
Expanding the determinant:
\[ = 1 \cdot \begin{vmatrix} -1 & 2 \\ -(x-3) & -1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 1 & 2 \\ x-2 & -1 \end{vmatrix} + 3 \cdot \begin{vmatrix} 1 & -1 \\ x-2 & -(x-3) \end{vmatrix} \]

- Calculate each of the 2x2 determinants:
- \( \begin{vmatrix} -1 & 2 \\ -(x-3) & -1 \end{vmatrix} = -1 \cdot -1 - 2 \cdot -(x-3) = 1 + 2(x-3) = 2x - 5 \)

- \( \begin{vmatrix} 1 & 2 \\ x-2 & -1 \end{vmatrix} = 1 \cdot -1 - 2 \cdot (x-2) = -1 - 2x + 4 = -2x + 3 \)

- \( \begin{vmatrix} 1 & -1 \\ x-2 & -(x-3) \end{vmatrix} = 1 \cdot -(x-3) - (-1)(x-2) = -x + 3 + x - 2 = 1 \)

- Substitute back into the determinant:
\[ 1(2x - 5) - 2(-2x + 3) + 3(1) = 0 \]
\[ 2x - 5 + 4x - 6 + 3 = 0 \]
\[ 6x - 8 = 0 \]

5. **Solve for \( x \)**:
\[ 6x = 8 \implies x = \frac{8}{6} = \frac{4}{3} \]

6. **Find \( \frac{1}{x} \)**:
\[ \frac{1}{x} = \frac{1}{\frac{4}{3}} = \frac{3}{4} \]

### Final Answer:
\[ \frac{1}{x} = \frac{3}{4} \]