The plane `x- y - z = 2` is rotated through an angle `pi/2` about its line of intersection with the plane `x + 2y + z = 2`.
Find its equation in the new position.

Answer

To find the equation of the plane after it has been rotated through an angle of \( \frac{\pi}{2} \) about its line of intersection with another plane, we follow these steps:

### Step 1: Identify the given planes
The equations of the two planes are:
1. Plane 1: \( x - y - z = 2 \)
2. Plane 2: \( x + 2y + z = 2 \)

### Step 2: Write the family of planes
The family of planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes, we get:
\[ (x - y - z - 2) + \lambda (x + 2y + z - 2) = 0 \]
This simplifies to:
\[ (1 + \lambda)x + (-1 + 2\lambda)y + (-1 + \lambda)z - (2 + 2\lambda) = 0 \]

### Step 3: Identify the normal vectors
The normal vector of Plane 1, \( \mathbf{n_1} \), is \( (1, -1, -1) \) and the normal vector of Plane 2, \( \mathbf{n_2} \), is \( (1, 2, 1) \).

### Step 4: Set up the condition for perpendicularity
Since the plane is rotated about the line of intersection, the normals of the planes must be perpendicular. Therefore, we set up the dot product:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = 0 \]
This gives:
\[ (1)(1) + (-1)(2) + (-1)(1) = 0 \]
Simplifying, we find:
\[ 1 - 2 - 1 = -2 \neq 0 \]
This means we need to find \( \lambda \) such that the normals are perpendicular.

### Step 5: Calculate the dot product
The normal vector of the family of planes is:
\[ \mathbf{n} = (1 + \lambda, -1 + 2\lambda, -1 + \lambda) \]
Setting the dot product with \( \mathbf{n_2} \) to zero:
\[ (1 + \lambda)(1) + (-1 + 2\lambda)(2) + (-1 + \lambda)(1) = 0 \]
Expanding this gives:
\[ 1 + \lambda - 2 + 4\lambda - 1 + \lambda = 0 \]
Combining like terms:
\[ 6\lambda - 2 = 0 \implies 6\lambda = 2 \implies \lambda = \frac{1}{3} \]

### Step 6: Substitute \( \lambda \) back into the family of planes
Substituting \( \lambda = \frac{1}{3} \) into the equation of the family of planes:
\[ \left(1 + \frac{1}{3}\right)x + \left(-1 + 2 \cdot \frac{1}{3}\right)y + \left(-1 + \frac{1}{3}\right)z - \left(2 + 2 \cdot \frac{1}{3}\right) = 0 \]
This simplifies to:
\[ \frac{4}{3}x + \left(-1 + \frac{2}{3}\right)y + \left(-1 + \frac{1}{3}\right)z - \left(2 + \frac{2}{3}\right) = 0 \]
Which further simplifies to:
\[ \frac{4}{3}x - \frac{1}{3}y - \frac{2}{3}z - \frac{8}{3} = 0 \]

### Step 7: Clear the fractions
Multiplying through by 3 to eliminate the fractions:
\[ 4x - y - 2z - 8 = 0 \]
Rearranging gives us the final equation of the new position of the plane:
\[ 4x - y - 2z = 8 \]

### Final Answer
The equation of the plane in its new position after rotation is:
\[ 4x - y - 2z = 8 \]