The vector equation of the plane through the point `hat(i)+2hat(j)-hat(k)` and perpendicular to the line of intersection of the plane `rcdot(3hat(i)-hat(j)+hat(k))=1 and rcdot(hat(i)+4hat(j)-2hat(k))=2`, is

Options

A. A. `rcdot(2hat(i)+hat(j)-13hat(k))=-1`
B. B. `rcdot(2hat(i)-7hat(j)-13hat(k))=1`
C. C. `rcdot(2hat(i)+7hat(j)+13hat(k))=0`
D. D. None of these

Answer

To find the vector equation of the plane that passes through the point \(\hat{i} + 2\hat{j} - \hat{k}\) and is perpendicular to the line of intersection of the two given planes, we can follow these steps:

### Step 1: Identify the normal vectors of the given planes
The equations of the two planes are:
1. \( \mathbf{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1 \)
2. \( \mathbf{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2 \)

From these equations, we can identify the normal vectors:
- For the first plane, the normal vector \( \mathbf{n_1} = 3\hat{i} - \hat{j} + \hat{k} \)
- For the second plane, the normal vector \( \mathbf{n_2} = \hat{i} + 4\hat{j} - 2\hat{k} \)

### Step 2: Find the direction vector of the line of intersection
The direction vector of the line of intersection of the two planes can be found by taking the cross product of the two normal vectors:
\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \]

Calculating the cross product:
\[ \mathbf{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{d} = \hat{i}((-1)(-2) - (1)(4)) - \hat{j}((3)(-2) - (1)(1)) + \hat{k}((3)(4) - (-1)(1)) \]
\[ = \hat{i}(2 - 4) - \hat{j}(-6 - 1) + \hat{k}(12 + 1) \]
\[ = -2\hat{i} + 7\hat{j} + 13\hat{k} \]

### Step 3: Formulate the equation of the required plane
The required plane is perpendicular to the direction vector \( \mathbf{d} \) and passes through the point \( \hat{i} + 2\hat{j} - \hat{k} \). The normal vector of the required plane can be taken as \( \mathbf{d} = -2\hat{i} + 7\hat{j} + 13\hat{k} \).

Using the point-normal form of the plane equation:
\[ \mathbf{r} \cdot \mathbf{n} = \mathbf{r_0} \cdot \mathbf{n} \]
where \( \mathbf{r_0} = \hat{i} + 2\hat{j} - \hat{k} \) and \( \mathbf{n} = -2\hat{i} + 7\hat{j} + 13\hat{k} \).

Calculating \( \mathbf{r_0} \cdot \mathbf{n} \):
\[ \mathbf{r_0} \cdot \mathbf{n} = (\hat{i} + 2\hat{j} - \hat{k}) \cdot (-2\hat{i} + 7\hat{j} + 13\hat{k}) \]
\[ = (1)(-2) + (2)(7) + (-1)(13) \]
\[ = -2 + 14 - 13 = -1 \]

Thus, the equation of the plane is:
\[ \mathbf{r} \cdot (-2\hat{i} + 7\hat{j} + 13\hat{k}) = -1 \]

### Final Answer
The vector equation of the plane is:
\[ \mathbf{r} \cdot (-2\hat{i} + 7\hat{j} + 13\hat{k}) = -1 \]