Consider a plane `prod:vecr*(2hati+hatj-hatk)=5`, a line `L_(1): vecr=(3hati-hatj+2hatk)+lambda(2hati-3hatj-hatk)` and a point `a(3, -4, 1)*L_(2)` is a line passing through A intersecting `L_(1)` and parallel to plane `prod`.
Q. Plane containing `L_(1) and L_(2)` is :

Options

A. parallel to yz-plane
B. parallel to x-axis
C. parallel to y-axis
D. passing through origin

Answer

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

### Step 1: Understand the given information
We have:
- A plane \( \pi: \vec{r} \cdot (2\hat{i} + \hat{j} - \hat{k}) = 5 \)
- A line \( L_1: \vec{r} = (3\hat{i} - \hat{j} + 2\hat{k}) + \lambda(2\hat{i} - 3\hat{j} - \hat{k}) \)
- A point \( A(3, -4, 1) \) through which line \( L_2 \) passes, and \( L_2 \) intersects \( L_1 \) and is parallel to the plane \( \pi \).

### Step 2: Identify the direction ratios of \( L_1 \)
The direction ratios of line \( L_1 \) can be extracted from its parametric form:
\[ \vec{d_1} = (2, -3, -1) \]

### Step 3: Set up the equation of the plane
The plane containing \( L_1 \) and \( L_2 \) can be expressed in the general form:
\[ l(x - 3) + m(y + 4) + n(z - 1) = 0 \]
where \( (l, m, n) \) are the direction ratios of the normal to the plane.

### Step 4: Use the condition of perpendicularity
Since the plane is perpendicular to line \( L_1 \), the dot product of the normal vector \( (l, m, n) \) and the direction ratios of \( L_1 \) must equal zero:
\[ 2l - 3m - n = 0 \quad \text{(Equation 1)} \]

### Step 5: Use the point \( A \) to find another equation
Substituting the coordinates of point \( A(3, -4, 1) \) into the plane equation gives:
\[ l(3 - 3) + m(-4 + 4) + n(1 - 1) = 0 \implies 0 = 0 \]
This does not provide new information.

### Step 6: Find the direction ratios of line \( L_2 \)
Since \( L_2 \) is parallel to the plane, its direction ratios must satisfy the plane's equation. Let the direction ratios of \( L_2 \) be \( (l, m, n) \).

### Step 7: Solve the equations
We have:
1. \( 2l - 3m - n = 0 \) (from perpendicularity)
2. The plane must also satisfy the condition of being parallel to the plane \( \pi \) given by \( 2x + y - z = 5 \).

### Step 8: Express \( n \) in terms of \( l \) and \( m \)
From Equation 1:
\[ n = 2l - 3m \]

### Step 9: Substitute into the plane equation
Substituting \( n \) into the plane equation gives:
\[ l(x - 3) + m(y + 4) + (2l - 3m)(z - 1) = 0 \]
Expanding this:
\[ lx - 3l + my + 4m + 2lz - 2l - 3mz + 3m = 0 \]
Combining like terms:
\[ lx + my + (2l - 3m)z + (-3l + 4m + 3m - 2l) = 0 \]

### Step 10: Finalize the equation of the plane
The final equation of the plane can be simplified and rearranged to:
\[ -y + 3z = 7 \]

### Conclusion
Thus, the equation of the plane containing \( L_1 \) and \( L_2 \) is:
\[ -y + 3z = 7 \]