Equation of a plane passing through the intersection of the planes `vecr.(3hati-hatj+hatk)=1` and `vecr.(hati+4hatj-2hatk)=2` and passing through the point `(hati+2hatj-hatk)` is :

Options

A. `vecr.(2hati+7hatj-13hatk)=1`
B. `vecr.(2hati-7hatj-13hatk)=1`
C. `vecr.(2hati+7hatj+13hatk)=0`
D. none of these

Explanation

To find the equation of the plane passing through the intersection of the planes given by \( \vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1 \) and \( \vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2 \), and also passing through the point \( (1, 2, -1) \), we can follow these steps:

### Step 1: Write the equations of the given planes in Cartesian form.
The equations of the planes can be rewritten as:
1. \( 3x - y + z - 1 = 0 \) (Plane 1)
2. \( x + 4y - 2z - 2 = 0 \) (Plane 2)

### Step 2: Form the general equation of the family of planes.
Any plane passing through the intersection of the two given planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes:
\[ (3x - y + z - 1) + \lambda (x + 4y - 2z - 2) = 0 \]

### Step 3: Expand and simplify the equation.
Expanding the equation:
\[ 3x - y + z - 1 + \lambda x + 4\lambda y - 2\lambda z - 2\lambda = 0 \]
Combining like terms:
\[ (3 + \lambda)x + (-1 + 4\lambda)y + (1 - 2\lambda)z - (1 + 2\lambda) = 0 \]

### Step 4: Substitute the point into the equation.
We need the plane to pass through the point \( (1, 2, -1) \). Substituting \( x = 1 \), \( y = 2 \), and \( z = -1 \):
\[ (3 + \lambda)(1) + (-1 + 4\lambda)(2) + (1 - 2\lambda)(-1) - (1 + 2\lambda) = 0 \]
This simplifies to:
\[ 3 + \lambda - 2 + 8\lambda - 1 + 2\lambda - 1 - 2\lambda = 0 \]
Combining terms gives:
\[ (3 + \lambda - 2 - 1 - 1) + (8\lambda + 2\lambda - 2\lambda) = 0 \]
\[ (3 - 4) + 9\lambda = 0 \]
\[ -1 + 9\lambda = 0 \]
Thus:
\[ 9\lambda = 1 \implies \lambda = \frac{1}{9} \]

### Step 5: Substitute \( \lambda \) back into the plane equation.
Now substituting \( \lambda = \frac{1}{9} \) back into the equation of the plane:
\[ (3 + \frac{1}{9})x + (-1 + 4 \cdot \frac{1}{9})y + (1 - 2 \cdot \frac{1}{9})z - (1 + 2 \cdot \frac{1}{9}) = 0 \]
This simplifies to:
\[ \left(\frac{27}{9} + \frac{1}{9}\right)x + \left(-1 + \frac{4}{9}\right)y + \left(1 - \frac{2}{9}\right)z - \left(1 + \frac{2}{9}\right) = 0 \]
\[ \frac{28}{9}x + \left(-\frac{5}{9}\right)y + \left(\frac{7}{9}\right)z - \frac{11}{9} = 0 \]

### Step 6: Clear the fractions.
Multiplying through by 9 to eliminate the denominators:
\[ 28x - 5y + 7z - 11 = 0 \]
Thus, the equation of the plane is:
\[ 28x - 5y + 7z = 11 \]

### Final Answer:
The equation of the plane is:
\[ 28x - 5y + 7z = 11 \]