Let the line `L` intersect the lines `x - 2 = - y = z - 1, 2(x + 1) = 2(y - 1) = z + 1` and be parallel to the line `frac{x - 2}{3} = frac{y - 1}{1} = frac{z - 2}{2}`. Then which of the following points lies on L?

Options

A. `(-1/3, -1, -1)`
B. `(-1/3, -1, 1)`
C. `(-1/3, 1, 1)`
D. `(-1/3, 1, -1)`

Answer

To solve the given problem, we need to find the line \( L \) that intersects the two given lines and is parallel to the third line. Let's break it down step by step.

### Step 1: Write the equations of the given lines in parametric form.

1. **Line 1**: The equation \( x - 2 = -y = z - 1 \) can be rewritten as:
\[ x = 2 + t, \quad y = -t, \quad z = 1 + t \]
where \( t \) is a parameter.

2. **Line 2**: The equation \( 2(x + 1) = 2(y - 1) = z + 1 \) can be simplified to:
\[ x = -1 + \frac{1}{2}s, \quad y = 1 + \frac{1}{2}s, \quad z = -1 + s \]
where \( s \) is another parameter.

### Step 2: Find the direction ratios of the lines.

- For **Line 1**, the direction ratios are \( (1, -1, 1) \).
- For **Line 2**, the direction ratios are \( \left(\frac{1}{2}, \frac{1}{2}, 1\right) \).

### Step 3: Find the direction ratios of the line \( L \).

Let the direction ratios of line \( L \) be \( (a, b, c) \). Since line \( L \) is parallel to the line given by \( \frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2} \), we have:
\[ (a, b, c) = (3, 1, 2) \]

### Step 4: Set up the equations for the intersection.

For line \( L \) to intersect both lines, we need to equate the coordinates from both lines.

From **Line 1**:
\[ x = 2 + t, \quad y = -t, \quad z = 1 + t \]

From **Line 2**:
\[ x = -1 + \frac{1}{2}s, \quad y = 1 + \frac{1}{2}s, \quad z = -1 + s \]

### Step 5: Equate the coordinates.

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

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

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

### Step 6: Solve the system of equations.

From equation (3):
\[ t = s - 2 \quad \text{(4)} \]

Substituting (4) into (1):
\[ s - 2 + \frac{1}{2}s = -3 \quad \Rightarrow \quad \frac{3}{2}s - 2 = -3 \quad \Rightarrow \quad \frac{3}{2}s = -1 \quad \Rightarrow \quad s = -\frac{2}{3} \]

Using (4) to find \( t \):
\[ t = -\frac{2}{3} - 2 = -\frac{2}{3} - \frac{6}{3} = -\frac{8}{3} \]

### Step 7: Find the coordinates of the intersection point.

Substituting \( t \) into the equations of **Line 1**:
\[ x = 2 - \frac{8}{3} = \frac{6}{3} - \frac{8}{3} = -\frac{2}{3} \]
\[ y = -\left(-\frac{8}{3}\right) = \frac{8}{3} \]
\[ z = 1 - \frac{8}{3} = \frac{3}{3} - \frac{8}{3} = -\frac{5}{3} \]

Thus, the intersection point is:
\[ \left(-\frac{2}{3}, \frac{8}{3}, -\frac{5}{3}\right) \]

### Step 8: Check which of the given points lies on line \( L \).

Now we need to check the provided options to see which point matches the coordinates we found.