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To solve this problem, we need to find the equation of a plane that contains the point \( (1, -2, -3) \) and the line of intersection of the two given planes:

1. Plane 1: \( -x + y + 2z = 16 \)
2. Plane 2: \( -x - 2y - 2z = -1 \)

### Step-by-step solution:

1. **Find the direction vector of the line of intersection:**

   The direction vector of the line of intersection of the two planes is given by the cross product of the normal vectors of the two planes.

   The normal vector to the first plane, \( -x + y + 2z = 16 \), is \( \mathbf{n_1} = (-1, 1, 2) \), and the normal vector to the second plane, \( -x - 2y - 2z = -1 \), is \( \mathbf{n_2} = (-1, -2, -2) \).

   We compute the cross product \( \mathbf{n_1} \times \mathbf{n_2} \).

2. **Find a point on the line of intersection:**

   To find a point on the line of intersection, solve the system of equations formed by the two planes. This gives a point common to both planes, which lies on their intersection line.

3. **Form the equation of the plane:**

   The equation of the plane passing through the point \( (1, -2, -3) \) and containing the line of intersection will have a normal vector orthogonal to both the direction of the line (found in step 1) and any vector connecting the given point to a point on the line (from step 2).

Let's go through these steps to get the solution. I'll start by finding the cross product of the normal vectors first.The direction vector of the line of intersection of the two planes is \( \mathbf{d} = (2, -4, 3) \).

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

1. \( -x + y + 2z = 16 \)
2. \( -x - 2y - 2z = -1 \)

I'll solve this system of equations next to find 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 solve this system of equations manually by substitution or elimination:

1. Start with the two plane equations:
   \[
   -x + y + 2z = 16
   \]
   \[
   -x - 2y - 2z = -1
   \]

2. Solve for one variable from one of the equations and substitute into the other to find a common point on the line of intersection.

If you'd like, I can walk you through the steps or help with another question. Let me know!

Here are five related questions:
1. How do you compute the cross product of two vectors in 3D?
2. How do you find the parametric equations of a line given two planes?
3. What is the geometric significance of the intersection of two planes?
4. How can we derive the normal vector to a plane given its equation?
5. How do we verify if a point lies on a given plane?

Tip: The cross product of the normal vectors of two planes gives the direction vector of their line of intersection.

Find a plane containing the point (1, -2, -3) and the line of intersection of the planes -x + y + 2z = 16 and -x - 2y - 2z = -1.

Learn how to find a plane passing through the point (1, -2, -3) and the line of intersection of the planes -x + y + 2z = 16 and -x - 2y - 2z = -1. This solution involves using vector cross product, finding direction vectors, and solving systems of equations. Suitable for advanced high school or undergraduate students.

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