If `veca=hati-2hatj+3hatk, vecb=2hati+3hatj-hatk` and `vecc=rhati+hatj+(2r-1)hatk` are three vectors such that `vecc` is parallel to the plane of `veca` and `vecb` then`r` is equal to,

Options

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

Explanation

To solve the problem, we need to find the value of \( r \) such that the vector \( \vec{c} \) is parallel to the plane formed by the vectors \( \vec{a} \) and \( \vec{b} \). This can be determined by ensuring that the dot product of \( \vec{c} \) with the cross product of \( \vec{a} \) and \( \vec{b} \) is zero.

### Step-by-Step Solution:

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

2. **Calculate the cross product \( \vec{a} \times \vec{b} \)**:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix} \]

Expanding this determinant:
\[ = \hat{i} \begin{vmatrix} -2 & 3 \\ 3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 2 & 3 \end{vmatrix} \]

Calculating the minors:
\[ = \hat{i}((-2)(-1) - (3)(3)) - \hat{j}((1)(-1) - (3)(2)) + \hat{k}((1)(3) - (-2)(2)) \]
\[ = \hat{i}(2 - 9) - \hat{j}(-1 - 6) + \hat{k}(3 + 4) \]
\[ = -7\hat{i} + 7\hat{j} + 7\hat{k} \]
Thus,
\[ \vec{a} \times \vec{b} = -7\hat{i} + 7\hat{j} + 7\hat{k} \]

3. **Set up the dot product \( \vec{c} \cdot (\vec{a} \times \vec{b}) = 0 \)**:
\[ \vec{c} \cdot (\vec{a} \times \vec{b}) = (r\hat{i} + \hat{j} + (2r - 1)\hat{k}) \cdot (-7\hat{i} + 7\hat{j} + 7\hat{k}) \]

Calculating the dot product:
\[ = r(-7) + 1(7) + (2r - 1)(7) \]
\[ = -7r + 7 + 14r - 7 \]
\[ = 7r \]

4. **Set the equation to zero**:
\[ 7r = 0 \]

5. **Solve for \( r \)**:
\[ r = 0 \]

### Final Answer:
Thus, the value of \( r \) is \( \boxed{0} \).