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To find the equation of a plane that passes through the points \((2, 2, 1)\) and \((-1, 1, -1)\) and is perpendicular to the plane \(2x - 3y + z = 3\), follow these steps:

1. **Determine the normal vector to the given plane:**
   The plane \(2x - 3y + z = 3\) has a normal vector \(\vec{n_1} = \langle 2, -3, 1 \rangle\).

2. **Find the direction vector of the line passing through the given points:**
   The direction vector \(\vec{d}\) from \((2, 2, 1)\) to \((-1, 1, -1)\) is:
   \[
   \vec{d} = \langle -1 - 2, 1 - 2, -1 - 1 \rangle = \langle -3, -1, -2 \rangle
   \]

3. **Determine a normal vector to the required plane:**
   The normal vector \(\vec{n}\) to the required plane is perpendicular to both \(\vec{n_1}\) and \(\vec{d}\). To find \(\vec{n}\), compute the cross product \(\vec{n_1} \times \vec{d}\):
   \[
   \vec{n} = \vec{n_1} \times \vec{d} = \begin{vmatrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   2 & -3 & 1 \\
   -3 & -1 & -2
   \end{vmatrix}
   \]
   \[
   \vec{n} = \mathbf{i}((-3)(-2) - (1)(-1)) - \mathbf{j}((2)(-2) - (1)(-3)) + \mathbf{k}((2)(-1) - (-3)(-3))
   \]
   \[
   \vec{n} = \mathbf{i}(6 + 1) - \mathbf{j}(-4 + 3) + \mathbf{k}(-2 - 9)
   \]
   \[
   \vec{n} = \mathbf{i}(7) - \mathbf{j}(-1) + \mathbf{k}(-11)
   \]
   \[
   \vec{n} = \langle 7, 1, -11 \rangle
   \]

4. **Find the equation of the plane:**
   The general equation of a plane is given by:
   \[
   a(x - x_0) + b(y - y_0) + c(z - z_0) = 0
   \]
   where \((x_0, y_0, z_0)\) is a point on the plane. We can use \((2, 2, 1)\).

   Substituting the normal vector \(\langle 7, 1, -11 \rangle\) and the point \((2, 2, 1)\), we get:
   \[
   7(x - 2) + 1(y - 2) - 11(z - 1) = 0
   \]
   \[
   7x - 14 + y - 2 - 11z + 11 = 0
   \]
   \[
   7x + y - 11z - 5 = 0
   \]

So, the equation of the plane is:
\[ 7x + y - 11z - 5 = 0 \]

Would you like any more details or have any questions?

Here are 8 related questions to expand on this information:
1. How do you compute the cross product of two vectors?
2. Why is the cross product used to find a normal vector to a plane?
3. Can the normal vector be directly used as the plane equation coefficients?
4. What is the geometric significance of the normal vector to a plane?
5. How can you verify if a point lies on a given plane?
6. What happens if the given points and the direction vector are collinear?
7. Can there be multiple planes passing through the same points and being perpendicular to the same plane?
8. How would the plane equation change if a different point on the plane was used?

**Tip:** When solving for planes in 3D space, always double-check the normal vector calculations, as errors in cross product computation can lead to incorrect plane equations.

Learn how to find the equation of a plane passing through specified points and perpendicular to another plane using vector cross product and plane equation formulas. This problem involves advanced concepts in vectors and plane geometry.

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