Consider the lines `L_1` and `L_2` given by
`L_1:(x-1)/2=(y-3)/1=(z-2)/2 ,L_2 : (x-2)/1=(y-2)/2=(z-3)/3`
A line `L_3` having direction ratios 1, –1, –2, intersects `L_1` and `L_2` at the point P and Q respectively. The thelength of the line segment PQ is

Options

A. `3sqrt2`
B. 4
C. `2sqrt6`
D. `4sqrt3`

Answer

To solve the problem, we need to find the points of intersection of the line \( L_3 \) with lines \( L_1 \) and \( L_2 \), and then calculate the distance between these two points.

### Step 1: Parametrize the lines \( L_1 \) and \( L_2 \)

The lines are given in symmetric form. We can convert them into parametric form.

For line \( L_1 \):
\[ \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2} = r_1 \]
This gives us:
\[ x = 2r_1 + 1, \quad y = r_1 + 3, \quad z = 2r_1 + 2 \]

For line \( L_2 \):
\[ \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3} = r_2 \]
This gives us:
\[ x = r_2 + 2, \quad y = 2r_2 + 2, \quad z = 3r_2 + 3 \]

### Step 2: Parametrize line \( L_3 \)

The direction ratios of line \( L_3 \) are given as \( (1, -1, -2) \). Thus, we can express line \( L_3 \) as:
\[ x = t + x_0, \quad y = -t + y_0, \quad z = -2t + z_0 \]
We need to find suitable values for \( x_0, y_0, z_0 \).

### Step 3: Find intersection point \( P \) on line \( L_1 \)

Setting the equations of \( L_3 \) equal to those of \( L_1 \):
\[ t + x_0 = 2r_1 + 1 \]
\[ -t + y_0 = r_1 + 3 \]
\[ -2t + z_0 = 2r_1 + 2 \]

### Step 4: Find intersection point \( Q \) on line \( L_2 \)

Setting the equations of \( L_3 \) equal to those of \( L_2 \):
\[ t + x_0 = r_2 + 2 \]
\[ -t + y_0 = 2r_2 + 2 \]
\[ -2t + z_0 = 3r_2 + 3 \]

### Step 5: Solve the system of equations

From the equations for \( P \):
1. \( t + x_0 = 2r_1 + 1 \) (1)
2. \( -t + y_0 = r_1 + 3 \) (2)
3. \( -2t + z_0 = 2r_1 + 2 \) (3)

From the equations for \( Q \):
4. \( t + x_0 = r_2 + 2 \) (4)
5. \( -t + y_0 = 2r_2 + 2 \) (5)
6. \( -2t + z_0 = 3r_2 + 3 \) (6)

### Step 6: Express \( r_1 \) and \( r_2 \) in terms of \( t \)

From equation (1) and (4):
\[ 2r_1 + 1 = r_2 + 2 \implies r_2 = 2r_1 - 1 \]

From equation (2) and (5):
\[ r_1 + 3 = 2r_2 + 2 \implies r_1 = 2r_2 - 1 \]

### Step 7: Substitute and solve for \( r_1 \) and \( r_2 \)

Substituting \( r_2 = 2r_1 - 1 \) into \( r_1 = 2r_2 - 1 \):
\[ r_1 = 2(2r_1 - 1) - 1 \implies r_1 = 4r_1 - 2 - 1 \implies 3r_1 = 3 \implies r_1 = 1 \]
Then, substituting back:
\[ r_2 = 2(1) - 1 = 1 \]

### Step 8: Find points \( P \) and \( Q \)

Substituting \( r_1 = 1 \) into the equations for \( P \):
\[ P = (2(1) + 1, 1 + 3, 2(1) + 2) = (3, 4, 4) \]

Substituting \( r_2 = 1 \) into the equations for \( Q \):
\[ Q = (1 + 2, 2(1) + 2, 3(1) + 3) = (3, 4, 6) \]

### Step 9: Calculate the distance \( PQ \)

Using the distance formula:
\[ PQ = \sqrt{(3 - 3)^2 + (4 - 4)^2 + (6 - 4)^2} = \sqrt{0 + 0 + 4} = \sqrt{4} = 2 \]

### Final Answer
The length of the line segment \( PQ \) is \( 2 \).