Find the equation of a plane passing through the intersection of the planes `vecr . (hati+3hatj-hatk) = 5` and `vecr.(2hati-hatj+hatk) = 3` and passes through the point `(2,1,-2)`.

Options

A. `vecr.(3hati+2hatj)=8`
B. `vecr.(2hati+3hatj)=8`
C. `vecr.(3hati+2hatj)+8=0`
D. none of these

Explanation

To find the equation of a plane passing through the intersection of the given planes and through the point (2, 1, -2), we can follow these steps:

### Step 1: Identify the equations of the given planes
The equations of the two planes are given as:
1. \( \vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5 \)
2. \( \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3 \)

### Step 2: Convert the equations to Cartesian form
The first plane can be written in Cartesian form as:
\[ x + 3y - z - 5 = 0 \]
The second plane can be written as:
\[ 2x - y + z - 3 = 0 \]

### Step 3: Write the general equation of the plane through the intersection
The general equation of a plane passing through the intersection of the two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes:
\[ (x + 3y - z - 5) + \lambda(2x - y + z - 3) = 0 \]

### Step 4: Expand and rearrange the equation
Expanding the equation gives:
\[ x + 3y - z - 5 + \lambda(2x - y + z - 3) = 0 \]
This simplifies to:
\[ (1 + 2\lambda)x + (3 - \lambda)y + (-1 + \lambda)z - (5 + 3\lambda) = 0 \]

### Step 5: Substitute the point (2, 1, -2)
Since the plane passes through the point (2, 1, -2), we substitute these coordinates into the equation:
\[ (1 + 2\lambda)(2) + (3 - \lambda)(1) + (-1 + \lambda)(-2) - (5 + 3\lambda) = 0 \]
This simplifies to:
\[ (2 + 4\lambda) + (3 - \lambda) + (2 - 2\lambda) - (5 + 3\lambda) = 0 \]
Combining like terms:
\[ 2 + 4\lambda + 3 - \lambda + 2 - 2\lambda - 5 - 3\lambda = 0 \]
This further simplifies to:
\[ 0 + (4\lambda - \lambda - 2\lambda - 3\lambda) = 0 \]
Thus:
\[ -2\lambda = 0 \implies \lambda = 1 \]

### Step 6: Substitute \(\lambda\) back into the plane equation
Now substituting \(\lambda = 1\) back into the equation of the plane:
\[ (1 + 2(1))x + (3 - 1)y + (-1 + 1)z - (5 + 3(1)) = 0 \]
This simplifies to:
\[ 3x + 2y - 8 = 0 \]

### Step 7: Final form of the equation
The equation can be expressed in vector form as:
\[ 3\hat{i} + 2\hat{j} \cdot \vec{r} - 8 = 0 \]

### Final Answer
The equation of the required plane is:
\[ 3x + 2y = 8 \]