Find the equation of plane containing the line `x+ y - z = 0 = 2x – y + z` and passing through the point `(1, 2, 1)`

Options

A. `3x+ 2y + z = 1`
B. `x - y + z = 0`
C. `x + y + z = 0`
D. `2x + 3y + z = 2`

Answer

To find the equation of the plane containing the line defined by the intersection of the two planes \(P_1: x + y - z = 0\) and \(P_2: 2x - y + z = 0\), and passing through the point \((1, 2, 1)\), we can follow these steps:

### Step 1: Write the equations of the given planes
The equations of the two planes are:
1. \(P_1: x + y - z = 0\)
2. \(P_2: 2x - y + z = 0\)

### Step 2: Formulate the equation of the required plane
The required plane \(P\) can be expressed as a linear combination of the two planes \(P_1\) and \(P_2\):
\[ P: P_1 + \lambda P_2 = 0 \]
Substituting the equations of \(P_1\) and \(P_2\):
\[ (x + y - z) + \lambda(2x - y + z) = 0 \]

### Step 3: Simplify the equation
Expanding the equation:
\[ x + y - z + \lambda(2x - y + z) = 0 \]
This can be rearranged as:
\[ (1 + 2\lambda)x + (1 - \lambda)y + (-1 + \lambda)z = 0 \]

### Step 4: Substitute the point into the plane equation
Since the plane passes through the point \((1, 2, 1)\), we substitute \(x = 1\), \(y = 2\), and \(z = 1\) into the equation:
\[ (1 + 2\lambda)(1) + (1 - \lambda)(2) + (-1 + \lambda)(1) = 0 \]
This leads to:
\[ 1 + 2\lambda + 2 - 2\lambda - 1 + \lambda = 0 \]
Simplifying this gives:
\[ 2 + \lambda = 0 \]

### Step 5: Solve for \(\lambda\)
From the equation \(2 + \lambda = 0\), we find:
\[ \lambda = -2 \]

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

### Final Result
The equation of the required plane is:
\[ \boxed{x - y + z = 0} \]