Let `L1 and L2` be two lines such that
`L_(2) : (x+1)/-3=(y-3)/2=(z+2)/1, L_(2) : x/1 = (y-7)/-3 = (z+7)/2`
The point of intersection of `L1 and L2` is

Options

A. (a) `(-3, 2, 1)`
B. (b) `(2, 1, -3)`
C. (c) `(1, -3, 2)`
D. (d) (0, 0, 0)

Answer

To find the point of intersection of the two lines \( L_1 \) and \( L_2 \), we will follow these steps:

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

For line \( L_1 \):
\[ \frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1} = \lambda \]
From this, we can express \( x, y, z \) in terms of \( \lambda \):
- \( x = -3\lambda - 1 \)
- \( y = 2\lambda + 3 \)
- \( z = \lambda - 2 \)

For line \( L_2 \):
\[ \frac{x}{1} = \frac{y - 7}{-3} = \frac{z + 7}{2} = \mu \]
From this, we can express \( x, y, z \) in terms of \( \mu \):
- \( x = \mu \)
- \( y = -3\mu + 7 \)
- \( z = 2\mu - 7 \)

### Step 2: Set the equations equal to each other

Now we have:
1. \( x = -3\lambda - 1 \) and \( x = \mu \)
2. \( y = 2\lambda + 3 \) and \( y = -3\mu + 7 \)
3. \( z = \lambda - 2 \) and \( z = 2\mu - 7 \)

### Step 3: Solve the equations

From the first equation:
\[ -3\lambda - 1 = \mu \tag{1} \]

From the second equation:
\[ 2\lambda + 3 = -3\mu + 7 \tag{2} \]

From the third equation:
\[ \lambda - 2 = 2\mu - 7 \tag{3} \]

### Step 4: Substitute and solve for \( \lambda \) and \( \mu \)

First, substitute \( \mu \) from equation (1) into equations (2) and (3).

Substituting into equation (2):
\[ 2\lambda + 3 = -3(-3\lambda - 1) + 7 \]
\[ 2\lambda + 3 = 9\lambda + 3 + 7 \]
\[ 2\lambda + 3 = 9\lambda + 10 \]
\[ 2\lambda - 9\lambda = 10 - 3 \]
\[ -7\lambda = 7 \implies \lambda = -1 \]

Now substitute \( \lambda = -1 \) back into equation (1) to find \( \mu \):
\[ \mu = -3(-1) - 1 = 3 - 1 = 2 \]

### Step 5: Find the coordinates of the intersection point

Now substitute \( \lambda = -1 \) into the equations for \( L_1 \):
- \( x = -3(-1) - 1 = 3 - 1 = 2 \)
- \( y = 2(-1) + 3 = -2 + 3 = 1 \)
- \( z = -1 - 2 = -3 \)

Thus, the point of intersection is \( (2, 1, -3) \).

### Step 6: Verify with \( L_2 \)

Now substitute \( \mu = 2 \) into the equations for \( L_2 \):
- \( x = 2 \)
- \( y = -3(2) + 7 = -6 + 7 = 1 \)
- \( z = 2(2) - 7 = 4 - 7 = -3 \)

Both lines give the same point of intersection.

### Conclusion

The point of intersection of lines \( L_1 \) and \( L_2 \) is:
\[ \boxed{(2, 1, -3)} \]