Consider the lines: `L_1:(x-2)/1=(y-1)/7=(z+2)/-5, L_2:x-4=y+3=-z` Then which of the following is/are correct ? (A) Point of intersection of `L_1 and L_2 is (1,-6,3)`

Options

A. Point of intersection of `L_(1) and L_(2)` is (1, -6, 3)
B. Equation of plane containing `L_(1) and L_(2)` is `x+2y+3z+2=0`
C. Acute angle between `L_(1) and L_(2)` is `cot^(-1)((13)/(15))`
D. Equation of plane containing `L_(1) and L_(2)` is `x+2y+2z+3=0`

Answer

To solve the problem, we need to find the point of intersection of the two lines \( L_1 \) and \( L_2 \) given in the vector form. Let's break down the steps systematically.

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

The lines are given as:
- \( L_1: \frac{x-2}{1} = \frac{y-1}{7} = \frac{z+2}{-5} = \lambda \)
- \( L_2: x-4 = y+3 = -z = \mu \)

From these equations, we can express the coordinates in terms of parameters \( \lambda \) and \( \mu \).

For line \( L_1 \):
- \( x = 1\lambda + 2 \)
- \( y = 7\lambda + 1 \)
- \( z = -5\lambda - 2 \)

For line \( L_2 \):
- \( x = 1\mu + 4 \)
- \( y = 1\mu - 3 \)
- \( z = -\mu \)

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

Since we are looking for the point of intersection, we set the coordinates from both lines equal to each other:

1. \( 1\lambda + 2 = 1\mu + 4 \) (Equation 1)
2. \( 7\lambda + 1 = 1\mu - 3 \) (Equation 2)
3. \( -5\lambda - 2 = -\mu \) (Equation 3)

### Step 3: Solve the equations

**From Equation 1:**
\[ \lambda + 2 = \mu + 4 \implies \mu = \lambda - 2 \quad \text{(i)} \]

**Substituting (i) into Equation 2:**
\[ 7\lambda + 1 = (\lambda - 2) - 3 \]
\[ 7\lambda + 1 = \lambda - 5 \]
\[ 7\lambda - \lambda = -5 - 1 \]
\[ 6\lambda = -6 \implies \lambda = -1 \]

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

### Step 4: Find the point of intersection

Now we can substitute \( \lambda = -1 \) into the parametric equations of \( L_1 \) to find the coordinates of the point of intersection:

- For \( x \):
\[ x = 1(-1) + 2 = 1 \]
- For \( y \):
\[ y = 7(-1) + 1 = -6 \]
- For \( z \):
\[ z = -5(-1) - 2 = 3 \]

Thus, the point of intersection is \( (1, -6, 3) \).

### Conclusion

The point of intersection of lines \( L_1 \) and \( L_2 \) is indeed \( (1, -6, 3) \), confirming that option (A) is correct.