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.
The value of `sin^(-1)sinlamda` is equal to

Options

A. `3`
B. `pi-3`
C. `4`
D. `pi-4`

Explanation

To solve the problem, we need to determine the value of \( \sin^{-1}(\sin \lambda) \) given that the lines represented by the equations lie in the same plane.

### Step-by-Step Solution:

1. **Identify the Direction Ratios and Points of the Lines:**
The equations of the lines are given as:
- Line 1: \( \frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{\lambda} \)
- Line 2: \( \frac{x-2}{3} = \frac{y-3}{2} = \frac{z-2}{3} \)

From these equations, we can extract the direction ratios and points:
- For Line 1:
- Direction ratios: \( (2, 3, \lambda) \)
- Point: \( (3, 2, 1) \)
- For Line 2:
- Direction ratios: \( (3, 2, 3) \)
- Point: \( (2, 3, 2) \)

2. **Set Up the Determinant Condition:**
For the two lines to lie in the same plane, the following determinant must equal zero:
\[ \begin{vmatrix} 3 - 2 & 2 - 3 & 1 - 2 \\ 2 & 3 & \lambda \\ 3 & 2 & 3 \end{vmatrix} = 0 \]

This simplifies to:
\[ \begin{vmatrix} 1 & -1 & -1 \\ 2 & 3 & \lambda \\ 3 & 2 & 3 \end{vmatrix} = 0 \]

3. **Calculate the Determinant:**
Expanding the determinant:
\[ 1 \cdot \begin{vmatrix} 3 & \lambda \\ 2 & 3 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 2 & \lambda \\ 3 & 3 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} \]
This gives:
\[ 1(9 - 2\lambda) + (6 - 3\lambda) - (4 - 9) = 0 \]
Simplifying:
\[ 9 - 2\lambda + 6 - 3\lambda + 5 = 0 \]
\[ 20 - 5\lambda = 0 \]

4. **Solve for \( \lambda \):**
Rearranging gives:
\[ 5\lambda = 20 \implies \lambda = 4 \]

5. **Find \( \sin^{-1}(\sin \lambda) \):**
Now, we need to find \( \sin^{-1}(\sin 4) \). Since \( 4 \) radians is greater than \( \pi \) (approximately \( 3.14 \)), we can express \( 4 \) in terms of \( \pi \):
\[ 4 = \pi + (4 - \pi) \quad \text{(where \( 4 - \pi \) is in the third quadrant)} \]
In the third quadrant, \( \sin \) is negative, thus:
\[ \sin(4) = -\sin(4 - \pi) \]
Therefore:
\[ \sin^{-1}(\sin 4) = \pi - 4 \]

### Final Answer:
The value of \( \sin^{-1}(\sin \lambda) \) is \( \pi - 4 \).