To solve the problem, we need to determine the value of \( \sin^{-1}(\sin \lambda) \) given that the two lines lie in the same plane. Let's break down the solution step by step.
### Step 1: Write the equations of the lines in parametric form
The two lines are given as:
1. \( \frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{\lambda} \)
2. \( \frac{x-2}{3} = \frac{y-3}{2} = \frac{z-2}{3} \)
From these equations, we can express the lines in parametric form:
- For the first line, let \( t \) be the parameter:
\[
x_1 = 3 + 2t, \quad y_1 = 2 + 3t, \quad z_1 = 1 + \lambda t
\]
- For the second line, let \( s \) be the parameter:
\[
x_2 = 2 + 3s, \quad y_2 = 3 + 2s, \quad z_2 = 2 + 3s
\]
### Step 2: Determine the direction ratios and points
From the parametric equations, we can identify:
- Point on the first line: \( (3, 2, 1) \)
- Direction ratios of the first line: \( (2, 3, \lambda) \)
- Point on the second line: \( (2, 3, 2) \)
- Direction ratios of the second line: \( (3, 2, 3) \)
### Step 3: Set up the determinant condition
Since the two lines lie in the same plane, the determinant formed by the vectors must equal zero:
\[
\begin{vmatrix}
x_1 - x_2 & y_1 - y_2 & z_1 - z_2 \\
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2
\end{vmatrix} = 0
\]
Substituting the values:
\[
\begin{vmatrix}
3 - 2 & 2 - 3 & 1 - 2 \\
2 & 3 & \lambda \\
3 & 2 & 3
\end{vmatrix} = 0
\]
This simplifies to:
\[
\begin{vmatrix}
1 & -1 & -1 \\
2 & 3 & \lambda \\
3 & 2 & 3
\end{vmatrix} = 0
\]
### Step 4: Calculate the determinant
Calculating the determinant:
\[
1 \cdot \begin{vmatrix}
3 & \lambda \\
2 & 3
\end{vmatrix} - (-1) \cdot \begin{vmatrix}
2 & \lambda \\
3 & 3
\end{vmatrix} - 1 \cdot \begin{vmatrix}
2 & 3 \\
3 & 2
\end{vmatrix}
\]
Calculating each of these:
1. \( 3 \cdot 3 - 2 \cdot \lambda = 9 - 2\lambda \)
2. \( 2 \cdot 3 - 3 \cdot \lambda = 6 - 3\lambda \)
3. \( 2 \cdot 2 - 3 \cdot 3 = 4 - 9 = -5 \)
Putting it all together:
\[
(9 - 2\lambda) + (6 - 3\lambda) - 5 = 0
\]
This simplifies to:
\[
10 - 5\lambda = 0 \quad \Rightarrow \quad \lambda = 2
\]
### Step 5: Find \( \sin^{-1}(\sin \lambda) \)
Now, we need to find \( \sin^{-1}(\sin 2) \). Since \( 2 \) is in the range of \( \sin^{-1} \) (which is \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \)), we have:
\[
\sin^{-1}(\sin 2) = 2
\]
### Final Answer
Thus, the value of \( \sin^{-1}(\sin \lambda) \) is:
\[
\sin^{-1}(\sin 2) = 2
\]
