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

### Step 2: Form the general equation of the plane through the line of intersection
The general equation of a plane that passes through the line of intersection of two planes can be expressed as:
\[ P_1 + \lambda P_2 = 0 \]
Substituting the equations of the planes into this expression gives:
\[ (x + y - 2z - 1) + \lambda (x + 3y - z - 4) = 0 \]

### Step 3: Rearrange the equation
Expanding and rearranging the equation:
\[ x + y - 2z - 1 + \lambda x + 3\lambda y - \lambda z - 4\lambda = 0 \]
Combining like terms, we have:
\[ (1 + \lambda)x + (1 + 3\lambda)y + (-2 - \lambda)z - (1 + 4\lambda) = 0 \]

### Step 4: Substitute the point \( P_0 = (1, 0, 1) \)
Since the plane must pass through the point \( P_0 \), we substitute \( x = 1 \), \( y = 0 \), and \( z = 1 \) into the 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 terms results in:
\[ -6\lambda - 2 = 0 \]

### Step 5: Solve for \( \lambda \)
Solving for \( \lambda \):
\[ -6\lambda = 2 \implies \lambda = -\frac{1}{3} \]

### Step 6: Substitute \( \lambda \) back into the plane equation
Now, substitute \( \lambda = -\frac{1}{3} \) back into the plane equation:
\[ (1 - \frac{1}{3})x + (1 - 1)y + (-2 + \frac{1}{3})z - (1 - \frac{4}{3}) = 0 \]
This simplifies to:
\[ \frac{2}{3}x + 0y - \frac{5}{3}z + \frac{1}{3} = 0 \]
Multiplying through by 3 to eliminate the fractions gives:
\[ 2x - 5z + 1 = 0 \]

### Step 7: Rearranging the equation
Rearranging gives the final equation of the plane:
\[ 2x + 0y - 5z = -1 \]

### Final Answer:
The equation of the required plane is:
\[ 2x - 5z = -1 \]