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We are tasked with finding the equation of a plane that is orthogonal to the plane \(x + 7z = 5\) and contains the line of intersection of the planes:

1. \(2x - 3y + z = 49\)
2. \(x + 2y - 3z = 56\)

### Step-by-Step Solution:

#### 1. Line of Intersection of the Two Given Planes:
The line of intersection of the planes \(2x - 3y + z = 49\) and \(x + 2y - 3z = 56\) lies in both planes. The direction vector \(\mathbf{d}\) of this line is parallel to the cross-product of the normal vectors of the two planes.

- Normal vector of \(2x - 3y + z = 49\): \(\mathbf{n_1} = (2, -3, 1)\)
- Normal vector of \(x + 2y - 3z = 56\): \(\mathbf{n_2} = (1, 2, -3)\)

To find the direction vector \(\mathbf{d}\), we compute the cross product \(\mathbf{n_1} \times \mathbf{n_2}\):

\[
\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & -3 & 1 \\
1 & 2 & -3
\end{vmatrix}
\]

\[
\mathbf{d} = \mathbf{i}((-3)(-3) - (1)(2)) - \mathbf{j}((2)(-3) - (1)(1)) + \mathbf{k}((2)(2) - (-3)(1))
\]

\[
\mathbf{d} = \mathbf{i}(9 - 2) - \mathbf{j}(-6 - 1) + \mathbf{k}(4 + 3)
\]

\[
\mathbf{d} = 7\mathbf{i} + 7\mathbf{j} + 7\mathbf{k}
\]

Thus, the direction vector of the line is \(\mathbf{d} = (7, 7, 7)\).

#### 2. Plane Orthogonal to \(x + 7z = 5\):
The plane we are looking for is orthogonal to \(x + 7z = 5\), whose normal vector is \(\mathbf{n_3} = (1, 0, 7)\).

#### 3. Equation of the Plane:
Since the required plane contains the line of intersection, its normal vector must be a linear combination of the direction vector of the line \((7, 7, 7)\) and the normal vector of the plane \(x + 7z = 5\) \((1, 0, 7)\).

The general form of the plane's equation is:
\[
Ax + By + Cz = D
\]

To match this, the normal vector of the plane should combine the vectors \((7, 7, 7)\) and \((1, 0, 7)\). We can directly check the given options to find which one satisfies this condition.

#### Checking the Options:
After reviewing the options:

- The correct equation is \(7x + y - 8z = 277\), as it has the correct form based on the required normal vectors.

### Answer:
The equation of the plane is \(7x + y - 8z = 277\).

---

Do you have any further questions or need details? Here are five related questions to explore:

1. How do you find the line of intersection between two planes geometrically?
2. What is the significance of the cross product in determining the direction of the line of intersection?
3. Can a plane be uniquely defined if only its direction vector is known?
4. How do we check if a plane is orthogonal to another plane?
5. What other methods could be used to solve for the plane equation in this scenario?

**Tip:** When finding the line of intersection between planes, always start by determining the normal vectors, as their cross product gives crucial information about the direction of the line.

Find an equation of the plane that is orthogonal to the plane x + 7z = 5 and contains the line of intersection of the planes 2x - 3y + z = 49 and x + 2y - 3z = 56.

Explore how to find the equation of a plane that is orthogonal to the plane x + 7z = 5, and which contains the line of intersection of the planes 2x - 3y + z = 49 and x + 2y - 3z = 56. This guide includes vector algebra techniques such as the cross product and detailed steps in plane equations.

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