If the position vectors of the points A and B are `3hati+hatj+2hatk` and `hati-2hatj-4hatk` respectively then the equation of the plane through B and perpendicular to AB is

Options

A. `vecr.(2hati+3hatj+6hatk)=28`
B. `vecr.(2hati+3hatj+6hatk)=32`
C. `hatr.(2hati+3hatj+6hatk)+28=0`
D. none of these

Explanation

To find the equation of the plane through point B and perpendicular to the line segment AB, we can follow these steps:

### Step 1: Identify the position vectors of points A and B
The position vectors are given as:
- \( \vec{A} = 3\hat{i} + \hat{j} + 2\hat{k} \)
- \( \vec{B} = \hat{i} - 2\hat{j} - 4\hat{k} \)

### Step 2: Calculate the vector AB
The vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = \vec{B} - \vec{A} = (\hat{i} - 2\hat{j} - 4\hat{k}) - (3\hat{i} + \hat{j} + 2\hat{k}) \]
\[ = (1 - 3)\hat{i} + (-2 - 1)\hat{j} + (-4 - 2)\hat{k} \]
\[ = -2\hat{i} - 3\hat{j} - 6\hat{k} \]

### Step 3: Identify the normal vector to the plane
The normal vector \( \vec{n} \) of the plane is the same as the vector \( \vec{AB} \):
\[ \vec{n} = -2\hat{i} - 3\hat{j} - 6\hat{k} \]

### Step 4: Use the point-normal form of the plane equation
The equation of a plane can be expressed in the form:
\[ \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \]
where \( \vec{a} \) is the position vector of point B, and \( \vec{n} \) is the normal vector.

### Step 5: Calculate \( \vec{a} \cdot \vec{n} \)
Substituting \( \vec{a} = \hat{i} - 2\hat{j} - 4\hat{k} \) and \( \vec{n} = -2\hat{i} - 3\hat{j} - 6\hat{k} \):
\[ \vec{a} \cdot \vec{n} = (\hat{i} - 2\hat{j} - 4\hat{k}) \cdot (-2\hat{i} - 3\hat{j} - 6\hat{k}) \]
\[ = (1)(-2) + (-2)(-3) + (-4)(-6) \]
\[ = -2 + 6 + 24 = 28 \]

### Step 6: Write the equation of the plane
Now substituting back into the plane equation:
\[ \vec{r} \cdot (-2\hat{i} - 3\hat{j} - 6\hat{k}) = 28 \]
This can be rearranged to:
\[ \vec{r} \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) + 28 = 0 \]

### Final Equation of the Plane
Thus, the equation of the required plane is:
\[ \vec{r} \cdot (2\hat{i} + 3\hat{j} + 6\hat{k}) + 28 = 0 \]