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To solve the problem, we need to find the equation of the plane that passes through the point \( P(1, 2, 2) \) and contains the line given by:

\[
\textbf{l} : x(t) = 2t, \; y(t) = 2t, \; z(t) = 2 + 3t
\]

### Step 1: Find a Point on the Line
The parametric equations of the line are:
\[
x(t) = 2t, \quad y(t) = 2t, \quad z(t) = 2 + 3t
\]

We can choose a specific value of \( t \) to find a point on the line. Let's take \( t = 0 \):
\[
P_1: (x(0), y(0), z(0)) = (0, 0, 2)
\]
So, \( P_1(0, 0, 2) \) is a point on the line.

### Step 2: Find the Direction Vector of the Line
The direction vector \( \mathbf{d} \) of the line can be determined from the coefficients of \( t \) in the parametric equations:
\[
\mathbf{d} = \langle 2, 2, 3 \rangle
\]

### Step 3: Find Two Vectors on the Plane
Now, we can find two vectors on the plane:
1. \( \mathbf{v_1} \) from the point \( P(1, 2, 2) \) to \( P_1(0, 0, 2) \):
\[
\mathbf{v_1} = \langle 0 - 1, 0 - 2, 2 - 2 \rangle = \langle -1, -2, 0 \rangle
\]
2. The direction vector \( \mathbf{v_2} = \mathbf{d} = \langle 2, 2, 3 \rangle \) of the line is also on the plane.

### Step 4: Find the Normal Vector to the Plane
The normal vector \( \mathbf{n} \) to the plane can be found by taking the cross product of \( \mathbf{v_1} \) and \( \mathbf{v_2} \):
\[
\mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -2 & 0 \\ 2 & 2 & 3 \end{vmatrix}
\]

Calculating the determinant:
\[
\mathbf{n} = \mathbf{i} \left((-2)(3) - (0)(2)\right) - \mathbf{j} \left((-1)(3) - (0)(2)\right) + \mathbf{k} \left((-1)(2) - (-2)(2)\right)
\]
\[
\mathbf{n} = \mathbf{i} (-6) - \mathbf{j} (-3) + \mathbf{k} (2)
\]
\[
\mathbf{n} = \langle -6, 3, 2 \rangle
\]

### Step 5: Equation of the Plane
The equation of the plane is:
\[
-6(x - 1) + 3(y - 2) + 2(z - 2) = 0
\]
Expanding and simplifying:
\[
-6x + 6 + 3y - 6 + 2z - 4 = 0
\]
\[
-6x + 3y + 2z - 4 = 0
\]
\[
6x - 3y - 2z + 4 = 0
\]
This matches the last option in the given choices, so the correct answer is **\( \boxed{6x - 3y - 2z + 4 = 0} \)**.

Would you like more details or have any questions?

### Related Questions:
1. How do you compute the cross product of two vectors?
2. What is the geometric interpretation of the normal vector to a plane?
3. Can you derive the general equation of a plane given three points?
4. How do you determine if a point lies on a plane?
5. What are the applications of vector cross product in physics?

### Tip:
Always verify the final equation by substituting the original points to check if they satisfy the equation of the plane.

Find an equation for the plane which passes through the point P(1, 2, 2) and contains the line: l : x(t) = 2t, y(t) = 2t, z(t) = 2 + 3t.

This detailed solution guides you through finding the equation of a plane that passes through a specific point and contains a given line. It covers key vector geometry concepts, including the cross product and plane equation. Suitable for students in Grades 11-12.

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