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 the equations of two lines in space. Let's break this down step-by-step.

### Step 1: Identify the points and direction vectors of the lines

The first line is given by the equation:
\[ \frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{\lambda} \]
From this, we can identify:
- A point \( P(3, 2, 1) \)
- A direction vector \( \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} + \lambda\mathbf{k} \)

The second line is given by:
\[ \frac{x-2}{3} = \frac{y-3}{2} = \frac{z-2}{3} \]
From this, we can identify:
- A point \( Q(2, 3, 2) \)
- A direction vector \( \mathbf{b} = 3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \)

### Step 2: Find the vector \( \mathbf{PQ} \)

The vector \( \mathbf{PQ} \) from point \( P \) to point \( Q \) is calculated as follows:
\[ \mathbf{PQ} = Q - P = (2 - 3)\mathbf{i} + (3 - 2)\mathbf{j} + (2 - 1)\mathbf{k} = -\mathbf{i} + \mathbf{j} + \mathbf{k} \]

### Step 3: Set up the condition for coplanarity

For the lines to lie in the same plane, the vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{PQ} \) must be coplanar. This can be checked using the scalar triple product, which must equal zero:
\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{PQ}) = 0 \]

### Step 4: Calculate the cross product \( \mathbf{b} \times \mathbf{PQ} \)

Calculating the cross product:
\[ \mathbf{b} = 3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \]
\[ \mathbf{PQ} = -\mathbf{i} + \mathbf{j} + \mathbf{k} \]
Using the determinant to find \( \mathbf{b} \times \mathbf{PQ} \):
\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & 3 \\ -1 & 1 & 1 \end{vmatrix} \]
Calculating this determinant:
\[ = \mathbf{i}(2 \cdot 1 - 3 \cdot 1) - \mathbf{j}(3 \cdot 1 - 3 \cdot (-1)) + \mathbf{k}(3 \cdot 1 - 2 \cdot (-1)) \]
\[ = \mathbf{i}(2 - 3) - \mathbf{j}(3 + 3) + \mathbf{k}(3 + 2) \]
\[ = -\mathbf{i} - 6\mathbf{j} + 5\mathbf{k} \]

### Step 5: Dot product with \( \mathbf{a} \)

Now, we compute the dot product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{PQ}) \):
\[ \mathbf{a} = 2\mathbf{i} + 3\mathbf{j} + \lambda\mathbf{k} \]
\[ \mathbf{b} \times \mathbf{PQ} = -\mathbf{i} - 6\mathbf{j} + 5\mathbf{k} \]
Calculating the dot product:
\[ = 2(-1) + 3(-6) + \lambda(5) \]
\[ = -2 - 18 + 5\lambda = 5\lambda - 20 \]

### Step 6: Set the scalar triple product to zero

Setting the scalar triple product to zero gives:
\[ 5\lambda - 20 = 0 \]
Solving for \( \lambda \):
\[ 5\lambda = 20 \implies \lambda = 4 \]

### Step 7: Calculate \( \sin^{-1}(\sin \lambda) \)

Now we need to find \( \sin^{-1}(\sin \lambda) \):
\[ \sin^{-1}(\sin 4) \]
Since \( 4 \) radians is greater than \( \frac{\pi}{2} \) and less than \( \frac{3\pi}{2} \), we can use the property:
\[ \sin^{-1}(\sin x) = x - 2\pi n \quad \text{for } n \in \mathbb{Z} \]
In our case, we can find the equivalent angle in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \):
\[ \sin(4) = \sin(\pi - 4) \quad \text{(since } \sin(\pi - x) = \sin x\text{)} \]
Thus:
\[ \sin^{-1}(\sin 4) = \pi - 4 \]

### Final Answer

The value of \( \sin^{-1}(\sin \lambda) \) is:
\[ \boxed{\pi - 4} \]