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The problem in the image asks you to prove that the equations of two lines can be transformed into symmetric form by shifting the origin to a suitable point.

Let's break down the task:

1. **Equations of Lines**:  
   - Line 1: \(\frac{x - 2}{3} = \frac{y - 3}{-2} = \frac{z - 1}{1}\)  
   - Line 2: \(\frac{x - 1}{2} = \frac{y - 2}{-1} = \frac{z - 3}{-2}\)

2. **Objective**:  
   We need to show that these equations can be converted into symmetric form by changing the origin to a suitable point.

### Solution:

To find the symmetric form of the line equations, we need to express them in terms of a common parameter. Then, determine the new origin such that both lines intersect at this point.

#### Steps:

1. **Parameterize the Lines**:
   - For Line 1:  
     \[
     x = 3t + 2, \quad y = -2t + 3, \quad z = t + 1
     \]
     Where \(t\) is a parameter.
     
   - For Line 2:  
     \[
     x = 2s + 1, \quad y = -s + 2, \quad z = -2s + 3
     \]
     Where \(s\) is a parameter.

2. **Find Intersection**:
   - Set the parametric equations equal to each other to find the point of intersection (new origin).

3. **Shifting Origin**:
   - Shift the origin to the intersection point and re-write the equations in symmetric form.

#### Intersection Point:

Setting the parametric equations equal:
- \(3t + 2 = 2s + 1\)  
- \(-2t + 3 = -s + 2\)  
- \(t + 1 = -2s + 3\)  

Solve these equations simultaneously for \(t\) and \(s\).

1. From \(3t + 2 = 2s + 1\):  
   \[
   3t = 2s - 1 \quad \Rightarrow \quad t = \frac{2s - 1}{3}
   \]

2. From \(-2t + 3 = -s + 2\):  
   \[
   -2t = -s - 1 \quad \Rightarrow \quad t = \frac{s + 1}{2}
   \]

Equating the two expressions for \(t\):  
\[
\frac{2s - 1}{3} = \frac{s + 1}{2}
\]

Cross-multiplying:  
\[
2(2s - 1) = 3(s + 1) \quad \Rightarrow \quad 4s - 2 = 3s + 3 \quad \Rightarrow \quad s = 5
\]

Substitute \(s = 5\) into one of the expressions for \(t\):  
\[
t = \frac{2(5) - 1}{3} = \frac{9}{3} = 3
\]

#### Intersection Point:
Using \(s = 5\) in Line 2's parametric equation:  
- \(x = 2(5) + 1 = 11\)
- \(y = -5 + 2 = -3\)
- \(z = -2(5) + 3 = -7\)

The new origin is \((11, -3, -7)\).

#### New Symmetric Equations:
1. For Line 1 with new origin: \(\frac{x - 11}{3} = \frac{y + 3}{-2} = \frac{z + 7}{1}\)
2. For Line 2 with new origin: \(\frac{x - 11}{2} = \frac{y + 3}{-1} = \frac{z + 7}{-2}\)

The lines are now in symmetric form with respect to the new origin.

Would you like to explore any details further or have any questions?

### Related Questions

1. What are the parametric equations of a line?
2. How do you determine the point of intersection between two lines in space?
3. What is the symmetric form of a line equation?
4. How can changing the origin affect the equation of a line?
5. What are the steps to shift the origin to a new point?
6. How can we convert parametric equations to symmetric form?
7. What is the significance of finding the intersection of two lines?
8. How do the direction ratios of a line relate to its parametric equations?

**Tip:** Symmetric equations can simplify the analysis of lines and their relationships in space, particularly for identifying intersections and calculating distances.

Explore how to convert line equations into symmetric form by shifting the origin to an intersection point. This advanced math problem involves parametric equations and symmetric form transformations, suitable for advanced high school students.

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