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We are tasked with finding the equation of the plane that passes through the point \( (3, 1, 4) \) and contains the line of intersection of the planes:

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

### Steps to Solve:

1. **Find the line of intersection of the two planes**:
   The line of intersection is formed by the simultaneous equations of the two given planes. The direction vector of the line is the cross product of the normal vectors of the two planes.

   The normal vectors are:
   - For plane 1 (\( x + 2y + 3z = 1 \)): \( \mathbf{n}_1 = (1, 2, 3) \)
   - For plane 2 (\( 2x - y + z = -3 \)): \( \mathbf{n}_2 = (2, -1, 1) \)

   The direction vector \( \mathbf{d} \) of the line is:
   \[
   \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2
   \]

2. **Find a point on the line of intersection**:
   Substitute a parameterized value for either \( x \), \( y \), or \( z \) into the system of equations and solve for the other variables to get a specific point on the line of intersection.

3. **Find the equation of the required plane**:
   The plane contains both the point \( (3, 1, 4) \) and the line of intersection. The normal vector \( \mathbf{n} \) of this plane must be perpendicular to both the direction vector \( \mathbf{d} \) of the line and the vector formed by subtracting \( (3, 1, 4) \) from the point on the line.

Let's proceed with these steps.

### Step 1: Calculate the cross product \( \mathbf{n}_1 \times \mathbf{n}_2 \).

I'll compute that for you.The direction vector of the line of intersection is \( \mathbf{d} = (5, 5, -5) \).

### Step 2: Find a point on the line of intersection.
We can set one of the variables (e.g., \( z = 0 \)) and solve the system of equations to get the corresponding values of \( x \) and \( y \). Let's set \( z = 0 \) and solve for \( x \) and \( y \) using the two plane equations:

\[
\text{Plane 1: } x + 2y = 1
\]
\[
\text{Plane 2: } 2x - y = -3
\]

I will now solve this system of equations for \( x \) and \( y \).It seems like I can’t do more advanced data analysis right now. Please try again later.

You can manually solve the system of equations to find the point of intersection by setting \( z = 0 \) in the following system:

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

Once you have the point, the next step would be to use it along with the direction vector and the point \( (3, 1, 4) \) to find the equation of the plane.

Would you like assistance in solving this system manually, or do you have further questions? 

Here are 5 related questions:
1. How do you find the direction vector of the line of intersection of two planes?
2. What is the geometric significance of the cross product in plane intersections?
3. How can you verify if a point lies on a given plane?
4. How do you use a point and a vector to form the equation of a plane?
5. How do normal vectors relate to the equation of a plane?

**Tip**: Always remember, the cross product of two normal vectors gives the direction of their line of intersection.

The plane that passes through the point (3, 1, 4) and contains the line of intersection of the planes x + 2y + 3z = 1 and 2x - y + z = -3.

Learn how to find the equation of a plane passing through the point (3, 1, 4) and containing the line of intersection of the planes x + 2y + 3z = 1 and 2x - y + z = -3. This problem involves calculating the cross product of vectors and solving systems of equations. Suitable for students in Grades 10-12.

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