Two line whose are `(x-3)/(2)=(y-2)/(3)=(z-1)/(lambda) and (x-2)/(3)=(y-3)/(2)=(z-2)/(3)` lie in the same plane, then,
Q. Angle between the plane containing both the lines and the plane `4x+y+2z=0` is equal to

Options

A. `(phi)/(3)`
B. `(phi)/(2)`
C. `(phi)/(6)`
D. `cos^(-1)((2)/(sqrt(186)))`

Answer

To solve the problem, we need to find the angle between the plane containing two given lines and the plane defined by the equation \(4x + y + 2z = 0\).

### Step-by-Step Solution:

1. **Identify the Lines**:
The two lines are given as:
\[ \frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{\lambda} \quad \text{(Line 1)} \]
\[ \frac{x-2}{3} = \frac{y-3}{2} = \frac{z-2}{3} \quad \text{(Line 2)} \]

2. **Extract Position and Direction Vectors**:
- For Line 1, the position vector \( \mathbf{a_1} = (3, 2, 1) \) and the direction vector \( \mathbf{b_1} = (2, 3, \lambda) \).
- For Line 2, the position vector \( \mathbf{a_2} = (2, 3, 2) \) and the direction vector \( \mathbf{b_2} = (3, 2, 3) \).

3. **Calculate \( \mathbf{a_2} - \mathbf{a_1} \)**:
\[ \mathbf{a_2} - \mathbf{a_1} = (2-3, 3-2, 2-1) = (-1, 1, 1) \]

4. **Set Up the Scalar Triple Product**:
The lines are coplanar if the scalar triple product \( (\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) = 0 \).

5. **Calculate the Cross Product \( \mathbf{b_1} \times \mathbf{b_2} \)**:
\[ \mathbf{b_1} = (2, 3, \lambda), \quad \mathbf{b_2} = (3, 2, 3) \]
The cross product is calculated using the determinant:
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & \lambda \\ 3 & 2 & 3 \end{vmatrix} \]
Expanding this determinant:
\[ = \mathbf{i}(3 \cdot 3 - 2 \cdot \lambda) - \mathbf{j}(2 \cdot 3 - 3 \cdot \lambda) + \mathbf{k}(2 \cdot 2 - 3 \cdot 3) \]
\[ = \mathbf{i}(9 - 2\lambda) - \mathbf{j}(6 - 3\lambda) + \mathbf{k}(4 - 9) \]
\[ = (9 - 2\lambda, -(6 - 3\lambda), -5) \]

6. **Set Up the Scalar Triple Product**:
\[ (-1, 1, 1) \cdot (9 - 2\lambda, -(6 - 3\lambda), -5) = 0 \]
This gives:
\[ -1(9 - 2\lambda) + 1(-(6 - 3\lambda)) + 1(-5) = 0 \]
Simplifying:
\[ -9 + 2\lambda - 6 + 3\lambda - 5 = 0 \]
\[ 5\lambda - 20 = 0 \implies \lambda = 4 \]

7. **Direction Vector of the Plane**:
The direction vector of the plane containing the lines is \( \mathbf{c_1} = (9 - 2\lambda, -(6 - 3\lambda), -5) \) with \( \lambda = 4 \):
\[ \mathbf{c_1} = (9 - 8, -6 + 12, -5) = (1, 6, -5) \]

8. **Normal Vector of the Given Plane**:
The normal vector of the plane \( 4x + y + 2z = 0 \) is \( \mathbf{c_2} = (4, 1, 2) \).

9. **Calculate the Angle Between the Two Planes**:
The cosine of the angle \( \theta \) between the two planes is given by:
\[ \cos \theta = \frac{\mathbf{c_1} \cdot \mathbf{c_2}}{|\mathbf{c_1}| |\mathbf{c_2}|} \]
First, calculate \( \mathbf{c_1} \cdot \mathbf{c_2} \):
\[ \mathbf{c_1} \cdot \mathbf{c_2} = 1 \cdot 4 + 6 \cdot 1 - 5 \cdot 2 = 4 + 6 - 10 = 0 \]
Since the dot product is zero, \( \cos \theta = 0 \) implies:
\[ \theta = 90^\circ \]

### Final Answer:
The angle between the plane containing both lines and the plane \( 4x + y + 2z = 0 \) is \( 90^\circ \).