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.