A line `L_(1)` with direction ratios `-3,2,4` passes through the point A(7,6,2) and a line `L_(2)` with directions ratios 2,1,3 passes through the point B(5,3,4). A line `L_(3)` with direction ratios `2,-2,-1` intersects `L_(1)` and `L_(3)` at C and D, resectively. The equation of the plane parallel to line `L_(1)` and containing line `L_(2)` is equal to

Options

A. `x+3y+4z=30`
B. `x+2y+z=15`
C. `2x-y+z=11`
D. `2x+17y-7z=33`

Explanation

To find the equation of the plane parallel to line \( L_1 \) and containing line \( L_2 \), we can follow these steps:

### Step 1: Identify the direction ratios of the lines
- The direction ratios for line \( L_1 \) are given as \( -3, 2, 4 \).
- The direction ratios for line \( L_2 \) are \( 2, 1, 3 \).

### Step 2: Use the point on line \( L_2 \)
- The point \( B(5, 3, 4) \) lies on line \( L_2 \).
- We will use this point to form the equation of the plane.

### Step 3: Write the general equation of the plane
The general equation of a plane can be written as:
\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \]
Substituting point \( B(5, 3, 4) \):
\[ a(x - 5) + b(y - 3) + c(z - 4) = 0 \]

### Step 4: Use the direction ratios of line \( L_1 \)
Since the plane is parallel to line \( L_1 \), the normal vector of the plane can be derived from the direction ratios of line \( L_1 \). Thus, we can write:
\[ -3a + 2b + 4c = 0 \]

### Step 5: Use the direction ratios of line \( L_2 \)
Similarly, since line \( L_2 \) lies in the plane, we can use its direction ratios:
\[ 2a + b + 3c = 0 \]

### Step 6: Solve the system of equations
We now have the following system of equations:
1. \( -3a + 2b + 4c = 0 \) (Equation 1)
2. \( 2a + b + 3c = 0 \) (Equation 2)

From Equation 2, we can express \( b \) in terms of \( a \) and \( c \):
\[ b = -2a - 3c \]

Substituting \( b \) into Equation 1:
\[ -3a + 2(-2a - 3c) + 4c = 0 \]
\[ -3a - 4a - 6c + 4c = 0 \]
\[ -7a - 2c = 0 \implies 7a = -2c \implies c = -\frac{7}{2}a \]

### Step 7: Substitute back to find \( b \)
Substituting \( c \) back into the expression for \( b \):
\[ b = -2a - 3\left(-\frac{7}{2}a\right) \]
\[ b = -2a + \frac{21}{2}a = \frac{17}{2}a \]

### Step 8: Choose a value for \( a \)
Let \( a = 2 \) (to simplify calculations):
- Then \( b = \frac{17}{2} \cdot 2 = 17 \)
- And \( c = -\frac{7}{2} \cdot 2 = -7 \)

### Step 9: Write the equation of the plane
Substituting \( a, b, c \) back into the plane equation:
\[ 2(x - 5) + 17(y - 3) - 7(z - 4) = 0 \]
Expanding this:
\[ 2x - 10 + 17y - 51 - 7z + 28 = 0 \]
\[ 2x + 17y - 7z - 33 = 0 \]

### Final Answer
The equation of the plane parallel to line \( L_1 \) and containing line \( L_2 \) is:
\[ 2x + 17y - 7z = 33 \]