Find the point at which the given lines intersect.
r	 = (2, 3, 0) + 	t(3, −3, 3) 
r	 = (5, 0, 3) + 	s(−3, 3, 0)

Find an equation of the plane that contains these lines.


### Part 1: Find the point of intersection of the given lines

We are given two parametric lines in vector form:

\[
\mathbf{r}_1 = (2, 3, 0) + t(3, -3, 3)
\]
\[
\mathbf{r}_2 = (5, 0, 3) + s(-3, 3, 0)
\]

The lines will intersect at a point if the coordinates are equal for some values of the parameters \( t \) and \( s \).

#### Express the components of the two lines:

For line 1:
\[
x_1 = 2 + 3t, \quad y_1 = 3 - 3t, \quad z_1 = 0 + 3t
\]

For line 2:
\[
x_2 = 5 - 3s, \quad y_2 = 0 + 3s, \quad z_2 = 3
\]

#### Set the coordinates equal to each other:

1. \( x_1 = x_2 \):
\[
2 + 3t = 5 - 3s
\]
\[
3t + 3s = 3 \quad \text{or} \quad t + s = 1 \quad \text{(Equation 1)}
\]

2. \( y_1 = y_2 \):
\[
3 - 3t = 3s
\]
\[
-3t - 3s = -3 \quad \text{or} \quad t + s = 1 \quad \text{(Equation 2, same as Equation 1)}
\]

3. \( z_1 = z_2 \):
\[
3t = 3
\]
\[
t = 1
\]

#### Solve for \( s \):
Using \( t = 1 \) in Equation 1:
\[
1 + s = 1 \quad \Rightarrow \quad s = 0
\]

Thus, the point of intersection occurs when \( t = 1 \) and \( s = 0 \).

#### Find the coordinates of the intersection point:
Substitute \( t = 1 \) into the parametric equation of the first line:
\[
x = 2 + 3(1) = 5, \quad y = 3 - 3(1) = 0, \quad z = 3(1) = 3
\]

So, the point of intersection is \( (5, 0, 3) \).

### Part 2: Find an equation of the plane containing the lines

To find the equation of the plane, we need:
1. A point on the plane (the intersection point \( (5, 0, 3) \)).
2. A normal vector to the plane.

We can find the normal vector by taking the cross product of the direction vectors of the two lines.

#### Direction vectors of the lines:
- Line 1: \( \mathbf{v}_1 = (3, -3, 3) \)
- Line 2: \( \mathbf{v}_2 = (-3, 3, 0) \)

#### Compute the cross product \( \mathbf{v}_1 \times \mathbf{v}_2 \):

\[
\mathbf{v}_1 \times \mathbf{v}_2 = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
3 & -3 & 3 \\
-3 & 3 & 0
\end{vmatrix}
\]

\[
= \mathbf{i} \left( (-3)(0) - (3)(3) \right) - \mathbf{j} \left( (3)(0) - (-3)(3) \right) + \mathbf{k} \left( (3)(3) - (-3)(-3) \right)
\]

\[
= \mathbf{i} (-9) - \mathbf{j} (9) + \mathbf{k} (9 - 9)
\]

\[
= -9\mathbf{i} - 9\mathbf{j} + 0\mathbf{k}
\]

Thus, the normal vector is \( \mathbf{n} = (-9, -9, 0) \), or simplified \( \mathbf{n} = (-1, -1, 0) \).

#### Equation of the plane:
The general equation of a plane is:
\[
Ax + By + Cz = D
\]
Substitute \( \mathbf{n} = (-1, -1, 0) \) and the point \( (5, 0, 3) \) into the equation:
\[
-1(x) - 1(y) + 0(z) = D
\]
Substitute \( (5, 0, 3) \) into the equation:
\[
-1(5) - 1(0) = D \quad \Rightarrow \quad D = -5
\]

Thus, the equation of the plane is:
\[
x + y = 5
\]

### Final Answers:
1. The point of intersection is \( (5, 0, 3) \).
2. The equation of the plane is \( x + y = 5 \).

Would you like more details or have any questions?

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Here are 5 related questions:
1. How can we verify if two lines are skew or parallel?
2. What is the geometric interpretation of the cross product?
3. How would you find the angle between two intersecting lines?
4. What is the method to find the distance between two skew lines?
5. Can two lines in 3D space have more than one intersection point?

**Tip:** The cross product is a powerful tool to find a vector perpendicular to two given vectors, commonly used in plane equations.

Find the point at which the given lines intersect. r = (2, 3, 0) + t(3, −3, 3), r = (5, 0, 3) + s(−3, 3, 0). Find an equation of the plane that contains these lines.

Find the intersection point of two parametric lines and derive the equation of a plane containing these lines. This problem covers vector algebra, cross product, and plane equations, providing a detailed solution for advanced high school or undergraduate students.

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