Lines whose equation are `(x-3)/2=(y-2)/3=(z-1)/(lamda)` and `(x-2)/3=(y-3)/2=(z-2)/3` lie in same plane, then.
Angle between the plane containing both lines and the plane `4x+y+2z=0` is

Options

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

Explanation

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

### Step 1: Write the equations of the lines in vector form.

The first line is given by:
\[ \frac{x - 3}{2} = \frac{y - 2}{3} = \frac{z - 1}{\lambda} \]
This can be expressed in vector form as:
\[ \vec{r_1} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 3 \\ \lambda \end{pmatrix} \]

The second line is given by:
\[ \frac{x - 2}{3} = \frac{y - 3}{2} = \frac{z - 2}{3} \]
This can be expressed in vector form as:
\[ \vec{r_2} = \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix} + s \begin{pmatrix} 3 \\ 2 \\ 3 \end{pmatrix} \]

### Step 2: Identify the direction vectors and points on the lines.

From the equations, we can identify:
- For line 1:
- Point \(A_1 = (3, 2, 1)\)
- Direction vector \(b_1 = (2, 3, \lambda)\)

- For line 2:
- Point \(A_2 = (2, 3, 2)\)
- Direction vector \(b_2 = (3, 2, 3)\)

### Step 3: Find the vector connecting the two points.

The vector \(A_2 - A_1\) is:
\[ A_2 - A_1 = \begin{pmatrix} 2 \\ 3 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \]

### Step 4: Use the condition for coplanarity.

For the lines to lie in the same plane, the scalar triple product of the vectors \(A_2 - A_1\), \(b_1\), and \(b_2\) must be zero:
\[ (A_2 - A_1) \cdot (b_1 \times b_2) = 0 \]

### Step 5: Calculate the cross product \(b_1 \times b_2\).

The cross product can be calculated using the determinant:
\[ b_1 \times b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & \lambda \\ 3 & 2 & 3 \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \begin{vmatrix} 3 & \lambda \\ 2 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & \lambda \\ 3 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} \]
\[ = \hat{i} (9 - 2\lambda) - \hat{j} (6 - 3\lambda) + \hat{k} (4 - 9) \]
\[ = (9 - 2\lambda) \hat{i} - (6 - 3\lambda) \hat{j} - 5 \hat{k} \]

### Step 6: Set up the equation for coplanarity.

Now we compute the dot product with \(A_2 - A_1\):
\[ \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 9 - 2\lambda \\ -(6 - 3\lambda) \\ -5 \end{pmatrix} = 0 \]
This gives:
\[ -1(9 - 2\lambda) + 1(-6 + 3\lambda) - 5 = 0 \]
\[ -9 + 2\lambda - 6 + 3\lambda - 5 = 0 \]
\[ 5\lambda - 20 = 0 \implies \lambda = 4 \]

### Step 7: Find the normal vector of the plane containing both lines.

The direction vectors are:
- \(b_1 = (2, 3, 4)\)
- \(b_2 = (3, 2, 3)\)

The normal vector \(n\) of the plane containing both lines is \(b_1 \times b_2\):
\[ b_1 \times b_2 = \begin{pmatrix} 9 - 8 \\ -6 + 12 \\ -5 \end{pmatrix} = (1, 6, -5) \]

### Step 8: Find the normal vector of the given plane.

The normal vector of the plane \(4x + y + 2z = 0\) is:
\[ n_2 = (4, 1, 2) \]

### Step 9: Find the angle between the two planes.

The cosine of the angle \(\theta\) between the two planes can be found using the dot product:
\[ \cos \theta = \frac{n_1 \cdot n_2}{|n_1| |n_2|} \]
Calculating the dot product:
\[ n_1 \cdot n_2 = (1)(4) + (6)(1) + (-5)(2) = 4 + 6 - 10 = 0 \]

Since the dot product is zero, the angle \(\theta\) is:
\[ \theta = \frac{\pi}{2} \]

### Final Answer:
The angle between the plane containing both lines and the plane \(4x + y + 2z = 0\) is \(\frac{\pi}{2}\) radians.

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