The vector parallel to the line of intersection of the planes `vecr.(3hati-hatj+hatk) = 1` and `vecr.(hati+4hatj-2hatk)=2` is :

Options

A. `-2hati+7hatj+13hatk`
B. `2hati+7hatj-13hatk`
C. `-2hati-7hatj+13hatk`
D. `2hati+7hatj+13hatk`

Explanation

To find the vector parallel to the line of intersection of the two given planes, we can follow these steps:

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

From these equations, we can extract the normal vectors:
- For Plane 1: \( \vec{n_1} = 3\hat{i} - \hat{j} + \hat{k} \)
- For Plane 2: \( \vec{n_2} = \hat{i} + 4\hat{j} - 2\hat{k} \)

### Step 2: Calculate the cross product of the normal vectors
The direction of the line of intersection of the two planes can be found by calculating the cross product of the normal vectors \( \vec{n_1} \) and \( \vec{n_2} \).

\[ \vec{v} = \vec{n_1} \times \vec{n_2} = (3\hat{i} - \hat{j} + \hat{k}) \times (\hat{i} + 4\hat{j} - 2\hat{k}) \]

### Step 3: Set up the determinant for the cross product
We can express the cross product using the determinant of a matrix:

\[ \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix} \]

### Step 4: Calculate the determinant
Calculating the determinant, we have:

\[ \vec{v} = \hat{i} \begin{vmatrix} -1 & 1 \\ 4 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} -1 & 1 \\ 4 & -2 \end{vmatrix} = (-1)(-2) - (1)(4) = 2 - 4 = -2 \]

2. For \( -\hat{j} \):
\[ \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} = (3)(-2) - (1)(1) = -6 - 1 = -7 \quad \text{(so it becomes } +7\hat{j}) \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} = (3)(4) - (-1)(1) = 12 + 1 = 13 \]

### Step 5: Combine the results
Putting it all together, we have:
\[ \vec{v} = -2\hat{i} + 7\hat{j} + 13\hat{k} \]

### Conclusion
Thus, the vector parallel to the line of intersection of the two planes is:
\[ \vec{v} = -2\hat{i} + 7\hat{j} + 13\hat{k} \]