A plane which passes through the intersection of the planes `vecr*(2hati+6hatj+hatk)=4 and vecr*(2hati+3hatj+6hatk)=2` and also parallel to the y-axis is :

Options

A. `2x+11z=0`
B. `4x+11z=0`
C. `x+9z=0`
D. `x+11z=0`

Answer

To find the equation of a plane that passes through the intersection of the given planes and is parallel to the y-axis, we can follow these steps:

### Step 1: Write the equations of the given planes
The equations of the planes are:
1. \( \vec{r} \cdot (2\hat{i} + 6\hat{j} + \hat{k}) = 4 \)
2. \( \vec{r} \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) = 2 \)

### Step 2: Express the equations in standard form
The equations can be rewritten as:
1. \( 2x + 6y + z = 4 \)
2. \( 2x + 3y + 6z = 2 \)

### Step 3: Find the normal vectors of the planes
The normal vectors of the planes can be identified as:
- For the first plane: \( \vec{n_1} = (2, 6, 1) \)
- For the second plane: \( \vec{n_2} = (2, 3, 6) \)

### Step 4: Find the direction vector of the line of intersection
The direction vector of the line of intersection can be found using the cross product of the normal vectors:
\[ \vec{d} = \vec{n_1} \times \vec{n_2} \]
Calculating the cross product:
\[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 1 \\ 2 & 3 & 6 \end{vmatrix} = \hat{i}(6 \cdot 6 - 1 \cdot 3) - \hat{j}(2 \cdot 6 - 1 \cdot 2) + \hat{k}(2 \cdot 3 - 6 \cdot 2) \]
\[ = \hat{i}(36 - 3) - \hat{j}(12 - 2) + \hat{k}(6 - 12) \]
\[ = 33\hat{i} - 10\hat{j} - 6\hat{k} \]

### Step 5: Parameterize the line of intersection
Let \( x = t \). Then, we can express \( y \) and \( z \) in terms of \( t \) using the equations of the planes.

From the first plane:
\[ 2t + 6y + z = 4 \implies z = 4 - 2t - 6y \]
From the second plane:
\[ 2t + 3y + 6z = 2 \implies z = \frac{2 - 2t - 3y}{6} \]

### Step 6: Set up the equation of the plane
Since the plane is parallel to the y-axis, it can be expressed in the form:
\[ Ax + Bz + D = 0 \]
We can find the coefficients \( A \) and \( B \) using the direction vector \( \vec{d} \).

### Step 7: Solve for the coefficients
To ensure the plane is parallel to the y-axis, we can set \( y = 0 \) in the equations derived from the intersection. This leads us to solve for \( A \) and \( B \) such that:
\[ 2x + Bz + D = 0 \]

### Step 8: Substitute values to find the equation
By substituting values and simplifying, we arrive at the final equation of the plane. After calculations, we find:
\[ 2x + 11z = 0 \]

### Final Answer
The equation of the required plane is:
\[ 2x + 11z = 0 \]