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The question asks to **select all the lines** that contain the point \((3, 2, 4)\) and are **parallel to the vector** \(\langle 1, 2, 3 \rangle\).

Let's break this down:

1. **General form of a parametric line equation**: 
   - The parametric equations of a line in 3D, passing through a point \((x_0, y_0, z_0)\) and parallel to a vector \(\langle a, b, c \rangle\), can be written as:
   \[
   x = x_0 + at, \quad y = y_0 + bt, \quad z = z_0 + ct
   \]
   Here, \( t \) is the parameter.

2. **Given information**:
   - The point on the line is \((3, 2, 4)\).
   - The direction vector of the line is \(\langle 1, 2, 3 \rangle\), which gives \(a = 1\), \(b = 2\), and \(c = 3\).

3. **Checking each line**:
   Now, let's check if the lines in the options satisfy both conditions:
   
   - **Option 1**: \(x = 1 + t\), \(y = -2 + 2t\), \(z = -2 + 3t\)
     - To check if it contains the point \((3, 2, 4)\), substitute \(x = 3\), \(y = 2\), and \(z = 4\) into the equations:
       \[
       1 + t = 3 \quad \Rightarrow \quad t = 2
       \]
       \[
       y = -2 + 2(2) = 2 \quad (\text{True})
       \]
       \[
       z = -2 + 3(2) = 4 \quad (\text{True})
       \]
     So, **this line contains the point** \((3, 2, 4)\), and the direction vector \(\langle 1, 2, 3 \rangle\) matches. **Option 1 is correct**.

   - **Option 2**: \(x = 2 - t\), \(y = 2\), \(z = 1 + 3t\)
     - Check if it contains \((3, 2, 4)\):
       \[
       2 - t = 3 \quad \Rightarrow \quad t = -1
       \]
       \[
       z = 1 + 3(-1) = -2 \quad (\text{False})
       \]
     This line does **not contain the point** \((3, 2, 4)\). **Option 2 is incorrect**.

   - **Option 3**: \(x = 3 + t\), \(y = 2 + 2t\), \(z = -4 + 3t\)
     - Check if it contains \((3, 2, 4)\):
       \[
       3 + t = 3 \quad \Rightarrow \quad t = 0
       \]
       \[
       y = 2 + 2(0) = 2 \quad (\text{True})
       \]
       \[
       z = -4 + 3(0) = -4 \quad (\text{False})
       \]
     This line does **not contain the point** \((3, 2, 4)\). **Option 3 is incorrect**.

   - **Option 4**: \(x = 4 + 2t\), \(y = 3 + 4t\), \(z = 5 + 6t\)
     - Check if it contains \((3, 2, 4)\):
       \[
       4 + 2t = 3 \quad \Rightarrow \quad t = -\frac{1}{2}
       \]
       \[
       y = 3 + 4(-\frac{1}{2}) = 1 \quad (\text{False})
       \]
     This line does **not contain the point** \((3, 2, 4)\). **Option 4 is incorrect**.

### Conclusion:
- **Option 1** is the only correct line that contains the point \((3, 2, 4)\) and is parallel to the vector \(\langle 1, 2, 3 \rangle\).

Would you like more details or further explanations?

Here are 5 related questions for further exploration:
1. How do you derive parametric equations for a line in 3D space?
2. How can we determine if two lines are parallel in 3D geometry?
3. What is the significance of the direction vector in line equations?
4. How would you convert parametric equations to vector form?
5. How do you find the intersection of two lines in 3D?

**Tip**: Always ensure the point satisfies all parametric equations for a line to confirm it lies on that line.

Select ALL the lines that contain the point (3, 2, 4) and is parallel to the vector (1, 2, 3).

Solve the problem of finding lines that pass through the point (3, 2, 4) and are parallel to the vector (1, 2, 3). Learn how to use parametric equations in 3D space and verify whether a point lies on a given line. This problem covers important geometry concepts in 3D suitable for high school students.

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