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The problem is asking us to find the equation of a plane that:

1. Is orthogonal to the plane \( 6x + 5y = 8 \).
2. Contains the line of intersection of the planes \( 2x - 3y + z = 42 \) and \( x + 2y - 3z = 49 \).

### Step-by-Step Solution:

1. **Direction Vector of the Line of Intersection:**
   - The line of intersection of two planes is parallel to a vector that is the cross product of their normal vectors.
   
   - For the plane \( 2x - 3y + z = 42 \), the normal vector is \( \mathbf{n}_1 = \langle 2, -3, 1 \rangle \).
   - For the plane \( x + 2y - 3z = 49 \), the normal vector is \( \mathbf{n}_2 = \langle 1, 2, -3 \rangle \).

   - The direction vector \( \mathbf{d} \) of the line of intersection is the cross product of \( \mathbf{n}_1 \) and \( \mathbf{n}_2 \):
     \[
     \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \langle 2, -3, 1 \rangle \times \langle 1, 2, -3 \rangle
     \]

     Let's calculate the cross product.

2. **Normal Vector of the Plane:**
   - The desired plane is orthogonal to the plane \( 6x + 5y = 8 \), so its normal vector must be parallel to the normal vector of this plane.
   - The normal vector of \( 6x + 5y = 8 \) is \( \mathbf{n}_3 = \langle 6, 5, 0 \rangle \).

3. **Finding the Equation of the Plane:**
   - The equation of a plane can be written as:
     \[
     A(x - x_0) + B(y - y_0) + C(z - z_0) = 0
     \]
     where \( \langle A, B, C \rangle \) is the normal vector, and \( (x_0, y_0, z_0) \) is a point on the plane.

   - Since the plane contains the line of intersection of the two given planes, we need a point on this line. We can find a point by solving the system of equations of the two planes \( 2x - 3y + z = 42 \) and \( x + 2y - 3z = 49 \).

Let's start by calculating the cross product and finding a point on the line of intersection. I will compute this now.The direction vector of the line of intersection is \( \mathbf{d} = \langle 7, 7, 7 \rangle \). This indicates that the line of intersection is parallel to the vector \( \langle 7, 7, 7 \rangle \).

### Next steps:
We now need to find a point on the line of intersection by solving the system of equations for the two planes:

1. \( 2x - 3y + z = 42 \)
2. \( x + 2y - 3z = 49 \)

Let me solve this system to get a point on the line.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can still manually solve the system of equations to find a point on the line of intersection by setting one variable (such as \( y = 0 \)) and solving for \( x \) and \( z \).

Would you like some more details or assistance with any part of this process? Here are five related questions that might be helpful:

1. How do you manually compute the cross product of two vectors?
2. What are the steps to find the equation of a plane given a normal vector and a point?
3. How can you solve a system of linear equations to find the line of intersection?
4. Why is the direction vector important when working with planes and lines?
5. How do you determine if two planes are parallel, orthogonal, or intersecting?

**Tip:** To solve a system of equations, a common strategy is to eliminate one variable by substitution or elimination techniques.

Find an equation of the plane that is orthogonal to the plane 6x + 5y = 8 and contains the line of intersection of the planes 2x - 3y + z = 42 and x + 2y - 3z = 49.

Learn how to find the equation of a plane that is orthogonal to 6x + 5y = 8 and contains the line of intersection between the planes 2x - 3y + z = 42 and x + 2y - 3z = 49. Explore the use of vectors, cross product, and system of equations to derive the solution.

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