Let 𝑃
be the plane passing through the points [2,0,0]
, [−3,3,−2]
, and [2,−1,1]
.

Where does the line [2,3,3]+𝑠[2,3,−1]
intersect this plane?

To find the intersection point of the line \(\mathbf{L}\) with the plane \(\mathbf{P}\), we can follow these steps:

1. **Determine the equation of the plane \(\mathbf{P}\).**
2. **Find the parameter \(s\) for which the line intersects the plane.**
3. **Substitute the parameter \(s\) back into the line equation to find the intersection point.**

Let's go through each step in detail.

### Step 1: Determine the Equation of the Plane

The plane \(\mathbf{P}\) passes through the points:
- \(A = [2, 0, 0]\)
- \(B = [-3, 3, -2]\)
- \(C = [2, -1, 1]\)

To find the equation of the plane, we need a normal vector \(\mathbf{n}\) to the plane. We can obtain this by calculating the cross product of two vectors lying on the plane.

#### Calculate vectors \(\mathbf{AB}\) and \(\mathbf{AC}\):

\[
\mathbf{AB} = B - A = [-3, 3, -2] - [2, 0, 0] = [-5, 3, -2]
\]

\[
\mathbf{AC} = C - A = [2, -1, 1] - [2, 0, 0] = [0, -1, 1]
\]

#### Compute the cross product \(\mathbf{n} = \mathbf{AB} \times \mathbf{AC}\):

\[
\mathbf{n} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-5 & 3 & -2 \\
0 & -1 & 1
\end{vmatrix}
= \mathbf{i}(3 \cdot 1 - (-2) \cdot (-1)) - \mathbf{j}(-5 \cdot 1 - (-2) \cdot 0) + \mathbf{k}(-5 \cdot (-1) - 3 \cdot 0)
\]

\[
= \mathbf{i}(3 - 2) - \mathbf{j}(-5) + \mathbf{k}(5)
\]

\[
= \mathbf{i}(1) + \mathbf{j}(5) + \mathbf{k}(5)
\]

\[
= [1, 5, 5]
\]

The normal vector \(\mathbf{n}\) is \([1, 5, 5]\).

#### Equation of the plane:

Using the point \(A = [2, 0, 0]\) and the normal vector \(\mathbf{n} = [1, 5, 5]\), the equation of the plane is:

\[
1(x - 2) + 5(y - 0) + 5(z - 0) = 0
\]

Simplifying, we get:

\[
x + 5y + 5z = 2
\]

### Step 2: Find the Parameter \(s\) for Line Intersection

The line \(\mathbf{L}\) is given by:

\[
\mathbf{r}(s) = [2, 3, 3] + s[2, 3, -1]
\]

This implies the parametric equations:

\[
x = 2 + 2s, \quad y = 3 + 3s, \quad z = 3 - s
\]

Substitute these into the plane equation \(x + 5y + 5z = 2\):

\[
(2 + 2s) + 5(3 + 3s) + 5(3 - s) = 2
\]

Simplify:

\[
2 + 2s + 15 + 15s + 15 - 5s = 2
\]

\[
2s + 15s - 5s + 2 + 15 + 15 = 2
\]

\[
12s + 32 = 2
\]

\[
12s = 2 - 32
\]

\[
12s = -30
\]

\[
s = -\frac{30}{12} = -\frac{5}{2}
\]

### Step 3: Find the Intersection Point

Substitute \(s = -\frac{5}{2}\) into the parametric equations of the line:

\[
x = 2 + 2\left(-\frac{5}{2}\right) = 2 - 5 = -3
\]

\[
y = 3 + 3\left(-\frac{5}{2}\right) = 3 - \frac{15}{2} = 3 - 7.5 = -4.5
\]

\[
z = 3 - \left(-\frac{5}{2}\right) = 3 + \frac{5}{2} = 3 + 2.5 = 5.5
\]

Thus, the intersection point is:

\[
\boxed{[-3, -4.5, 5.5]}
\]

### Verification:

To ensure accuracy, we verify if this point satisfies the plane equation:

\[
x + 5y + 5z = 2 \Rightarrow (-3) + 5(-4.5) + 5(5.5) = 2
\]

\[
-3 - 22.5 + 27.5 = 2
\]

\[
2 = 2
\]

The calculations are verified as correct, confirming that the intersection point is indeed \(\boxed{[-3, -4.5, 5.5]}\).

---

Feel free to ask if you have any questions or need further clarification! Here are some questions you might find useful:

1. **Can you explain the concept of a normal vector again?**
2. **Why do we use the cross product to find the normal vector?**
3. **How would the process change if we were given different points?**
4. **What if the line was parallel to the plane?**
5. **Is there a quicker method for solving plane intersection problems?**

### Tip:
When checking your work in problems like this, substitute back into both the plane and line equations to ensure consistency and correctness.

Explore how to find the intersection point of a line and a plane in 3D geometry. This problem involves using vectors, parametric equations, and plane geometry concepts. Suitable for advanced high school or college-level students.

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