Find an equation for the plane through `P_(0)=(1, 0, 1)` and passing through the line of intersection
of the planes `x +y - 2z = 1 and x + 3y - z = 4`.

Answer

To find the equation of the plane that passes through the point \( P_0 = (1, 0, 1) \) and the line of intersection of the planes given by the equations \( P_1: x + y - 2z = 1 \) and \( P_2: x + 3y - z = 4 \), we can follow these steps:

### Step 1: Write the equations of the planes
The equations of the two planes are:
1. \( P_1: x + y - 2z - 1 = 0 \)
2. \( P_2: x + 3y - z - 4 = 0 \)

### Step 2: Find the equation of the plane through the intersection of the two planes
The equation of the plane that passes through the line of intersection of two planes can be expressed as:
\[ P = P_1 + \lambda P_2 = 0 \]
This can be written as:
\[ (x + y - 2z - 1) + \lambda (x + 3y - z - 4) = 0 \]

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

### Step 4: Substitute the point \( P_0 \)
We need the plane to pass through the point \( P_0(1, 0, 1) \). Substitute \( x = 1 \), \( y = 0 \), and \( z = 1 \) into the plane equation:
\[ (1 + \lambda)(1) + (1 + 3\lambda)(0) + (-2 - \lambda)(1) - (1 + 4\lambda) = 0 \]
This simplifies to:
\[ 1 + \lambda - 2 - \lambda - 1 - 4\lambda = 0 \]
Combining like terms:
\[ -2\lambda - 2 = 0 \]

### Step 5: Solve for \( \lambda \)
Solving for \( \lambda \):
\[ -2\lambda = 2 \implies \lambda = -1 \]

### Step 6: Substitute \( \lambda \) back into the plane equation
Substituting \( \lambda = -1 \) back into the plane equation:
\[ (1 - 1)x + (1 - 3)y + (-2 + 1)z - (1 - 4) = 0 \]
This simplifies to:
\[ 0x - 2y - z + 3 = 0 \]
or:
\[ -2y - z + 3 = 0 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ 2y + z = 3 \]

### Final Equation of the Plane
Thus, the equation of the required plane is:
\[ 2y + z - 3 = 0 \]