Find an equation of the plane.
The plane that passes through the line of intersection of the planes 
x − z = 1 and y + 3z = 1
 and is perpendicular to the plane 
x + y − 4z = 2

To find the equation of the plane that passes through the line of intersection of the planes \(x - z = 1\) and \(y + 3z = 1\), and is also perpendicular to the plane \(x + y - 4z = 2\), we will go step by step.

### Step 1: Parametrize the family of planes through the line of intersection
The equation of any plane passing through the line of intersection of two planes \(x - z = 1\) and \(y + 3z = 1\) can be written as a linear combination of their equations:
\[
(x - z - 1) + \lambda(y + 3z - 1) = 0
\]
where \(\lambda\) is a parameter. Expanding this:
\[
x - z - 1 + \lambda(y + 3z - 1) = 0
\]
\[
x - z - 1 + \lambda y + 3\lambda z - \lambda = 0
\]
\[
x + \lambda y + (-1 + 3\lambda)z - (1 + \lambda) = 0
\]
Thus, the general equation of a plane through the line of intersection is:
\[
x + \lambda y + (3\lambda - 1)z = 1 + \lambda
\]

### Step 2: Apply the condition of perpendicularity
The plane we are looking for is also perpendicular to the plane \(x + y - 4z = 2\). The normal vector to this plane is \(\mathbf{n_1} = (1, 1, -4)\). For two planes to be perpendicular, their normal vectors must be orthogonal. The normal vector to the general plane \(x + \lambda y + (3\lambda - 1)z = 1 + \lambda\) is \(\mathbf{n_2} = (1, \lambda, 3\lambda - 1)\).

The condition of orthogonality between \(\mathbf{n_1}\) and \(\mathbf{n_2}\) is given by the dot product:
\[
\mathbf{n_1} \cdot \mathbf{n_2} = 0
\]
\[
(1)(1) + (1)(\lambda) + (-4)(3\lambda - 1) = 0
\]
Expanding this:
\[
1 + \lambda - 12\lambda + 4 = 0
\]
\[
5 - 11\lambda = 0
\]
\[
\lambda = \frac{5}{11}
\]

### Step 3: Write the equation of the desired plane
Substituting \(\lambda = \frac{5}{11}\) into the general equation of the plane:
\[
x + \frac{5}{11} y + \left(3\left(\frac{5}{11}\right) - 1\right)z = 1 + \frac{5}{11}
\]
Simplifying the coefficients:
\[
x + \frac{5}{11} y + \left(\frac{15}{11} - 1\right)z = \frac{16}{11}
\]
\[
x + \frac{5}{11} y + \frac{4}{11}z = \frac{16}{11}
\]
Multiplying through by 11 to clear the denominators:
\[
11x + 5y + 4z = 16
\]

Thus, the equation of the desired plane is:
\[
11x + 5y + 4z = 16
\]

### Summary:
The equation of the plane that passes through the line of intersection of the planes \(x - z = 1\) and \(y + 3z = 1\), and is perpendicular to the plane \(x + y - 4z = 2\), is:
\[
11x + 5y + 4z = 16
\]

Do you need further clarification or details?

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Here are five related questions to expand on this topic:

1. How do you find the line of intersection between two planes geometrically?
2. What does it mean for two planes to be perpendicular in terms of their normal vectors?
3. How can you find the angle between two planes?
4. What is the geometric interpretation of a plane passing through a line of intersection?
5. How can the condition of parallelism between planes be mathematically expressed?

**Tip:** When finding a plane through the intersection of two other planes, always use the parametric method by combining the plane equations with a parameter.

Find an equation of the plane. The plane that passes through the line of intersection of the planes x − z = 1 and y + 3z = 1 and is perpendicular to the plane x + y − 4z = 2.

Find the equation of a plane passing through the line of intersection of the planes x - z = 1 and y + 3z = 1, while also being perpendicular to the plane x + y - 4z = 2. This solution involves the use of vector algebra, parametric equations, and the condition of orthogonality, making it suitable for students at the high school or undergraduate level.

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