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We are tasked with showing that two lines are parallel and finding the equation of the plane they determine. The parametric equations of the lines are:

1. Line 1:
   \[
   x = -2 + t, \quad y = 3 + 2t, \quad z = 4 - t
   \]
2. Line 2:
   \[
   x = 3 - t, \quad y = 4 - 2t, \quad z = t
   \]

### Step 1: Check if the lines are parallel

The direction vectors of the lines can be extracted from their parametric forms. 

- For Line 1, the direction vector \( \mathbf{v}_1 \) is:
  \[
  \mathbf{v}_1 = \langle 1, 2, -1 \rangle
  \]

- For Line 2, the direction vector \( \mathbf{v}_2 \) is:
  \[
  \mathbf{v}_2 = \langle -1, -2, 1 \rangle
  \]

To determine if the lines are parallel, we check if the direction vectors are scalar multiples of each other. We see that:

\[
\mathbf{v}_2 = -\mathbf{v}_1
\]

This means that the lines are parallel since their direction vectors are negatives of each other (i.e., they are scalar multiples).

### Step 2: Find the equation of the plane

Even though the lines are parallel, they determine a plane. To find the equation of the plane, we need:
- A point on the plane.
- The normal vector to the plane.

#### Find a point on the plane:
We can use one point from each line. For instance, let \( t = 0 \) in both parametric equations:

- For Line 1 at \( t = 0 \):
  \[
  P_1 = (-2, 3, 4)
  \]
  
- For Line 2 at \( t = 0 \):
  \[
  P_2 = (3, 4, 0)
  \]

#### Find the normal vector to the plane:
The normal vector to the plane can be found using the cross product of the direction vector of one line and the vector formed by the two points \( P_1 \) and \( P_2 \).

- The vector \( \mathbf{P_1P_2} \) is:
  \[
  \mathbf{P_1P_2} = P_2 - P_1 = \langle 3 - (-2), 4 - 3, 0 - 4 \rangle = \langle 5, 1, -4 \rangle
  \]

Now, compute the cross product \( \mathbf{v}_1 \times \mathbf{P_1P_2} \):

\[
\mathbf{v}_1 \times \mathbf{P_1P_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 5 & 1 & -4 \end{vmatrix}
\]

The cross product determinant is:
\[
\mathbf{i}(2(-4) - (-1)(1)) - \mathbf{j}(1(-4) - (-1)(5)) + \mathbf{k}(1(1) - 2(5))
\]
\[
= \mathbf{i}(-8 + 1) - \mathbf{j}(-4 + 5) + \mathbf{k}(1 - 10)
\]
\[
= \mathbf{i}(-7) - \mathbf{j}(1) + \mathbf{k}(-9)
\]
\[
= \langle -7, -1, -9 \rangle
\]

Thus, the normal vector to the plane is \( \langle -7, -1, -9 \rangle \).

#### Equation of the plane:
The equation of a plane is given by:
\[
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0
\]
where \( \langle a, b, c \rangle \) is the normal vector, and \( (x_0, y_0, z_0) \) is a point on the plane.

Using the normal vector \( \langle -7, -1, -9 \rangle \) and the point \( P_1 = (-2, 3, 4) \):

\[
-7(x + 2) - 1(y - 3) - 9(z - 4) = 0
\]
Expanding this:
\[
-7x - 14 - y + 3 - 9z + 36 = 0
\]
\[
-7x - y - 9z + 25 = 0
\]

Thus, the equation of the plane is:
\[
7x + y + 9z = 25
\]

### Final Answer:
1. The lines are parallel.
2. The equation of the plane they determine is:
   \[
   7x + y + 9z = 25
   \]

Would you like further clarification or more details on any of the steps?

Here are some related questions to deepen your understanding:
1. How do you find the equation of a plane given three points?
2. What are the conditions for two lines to be skew, parallel, or intersecting?
3. How is the cross product of two vectors defined geometrically?
4. Can two parallel lines always determine a unique plane?
5. How do you verify if a point lies on a given plane?

**Tip:** The cross product of two vectors gives a vector that is perpendicular to both, making it useful in finding normal vectors to planes.

Show that the lines x = -2 + t, y = 3 + 2t, z = 4 - t and x = 3 - t, y = 4 - 2t, z = t are parallel and find an equation of the plane they determine.

This solution shows how to verify that two lines in 3D are parallel and how to find the equation of the plane they determine. It involves vector algebra concepts like direction vectors and the cross product. Suitable for undergraduate or advanced high school students.

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