A line from the origin meets the lines `(x-2)/1=(y-1)/-2=(z+1)/1` and `(x-8/3)/2=(y+3)/-1=(z-1)/1` at `P` and `Q` respectively. If length `PQ = d`, then `d^2` is

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To solve the problem, we need to find the coordinates of points P and Q where the line from the origin intersects the two given lines. Then we will calculate the distance \( PQ \) and find \( d^2 \).

### Step-by-Step Solution:

1. **Equation of the Line from the Origin**:
Let the equation of the line passing through the origin be given by:
\[ \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \]

2. **Parameterizing the Given Lines**:
The first line is given by:
\[ \frac{x-2}{1} = \frac{y-1}{-2} = \frac{z+1}{1} \]
Let \( t_1 \) be the parameter for this line. Then we can express the coordinates of any point on this line as:
\[ x_1 = 2 + t_1, \quad y_1 = 1 - 2t_1, \quad z_1 = -1 + t_1 \]

The second line is given by:
\[ \frac{x - \frac{8}{3}}{2} = \frac{y + 3}{-1} = \frac{z - 1}{1} \]
Let \( t_2 \) be the parameter for this line. Then we can express the coordinates of any point on this line as:
\[ x_2 = \frac{8}{3} + 2t_2, \quad y_2 = -3 - t_2, \quad z_2 = 1 + t_2 \]

3. **Finding the Intersection Points**:
The line from the origin can be parameterized as:
\[ x = ar, \quad y = br, \quad z = cr \]
We need to find \( r \) for the first line:
\[ ar = 2 + t_1, \quad br = 1 - 2t_1, \quad cr = -1 + t_1 \]
From these equations, we can express \( t_1 \) in terms of \( r \):
- From \( ar = 2 + t_1 \), we have \( t_1 = ar - 2 \).
- Substitute \( t_1 \) into the other equations to find \( b \) and \( c \).

Similarly, for the second line:
\[ ar = \frac{8}{3} + 2t_2, \quad br = -3 - t_2, \quad cr = 1 + t_2 \]
Again, express \( t_2 \) in terms of \( r \) and substitute.

4. **Setting Up the System of Equations**:
We will have a system of equations from the above substitutions. Solve these equations to find \( a, b, c \).

5. **Finding Coordinates of Points P and Q**:
After solving the equations, we will find the coordinates of points \( P \) and \( Q \):
- Let’s say \( P = (x_P, y_P, z_P) \) and \( Q = (x_Q, y_Q, z_Q) \).

6. **Calculating the Distance \( PQ \)**:
The distance \( PQ \) can be calculated using the distance formula:
\[ PQ = \sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2} \]

7. **Finding \( d^2 \)**:
Finally, we need to find \( d^2 \):
\[ d^2 = (x_Q - x_P)^2 + (y_Q - y_P)^2 + (z_Q - z_P)^2 \]

### Final Answer:
After performing the calculations, we find that \( d^2 = 6 \).