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. Point of intersection of the lines lies on

Options

A. `3x+y+z=20`
B. `2x+y+z=25`
C. `3x+2y+z=24`
D. `x=y=z`

Answer

To solve the problem, we need to find the point of intersection of the two lines given by the equations:

1. \((x-3)/2 = (y-2)/3 = (z-1)/\lambda\)
2. \((x-2)/3 = (y-3)/2 = (z-2)/3\)

### Step 1: Identify Direction Ratios and Points on the Lines
From the first line, we can identify the direction ratios and a point on the line:
- Direction ratios: \( (2, 3, \lambda) \)
- Point on the line: \( (3, 2, 1) \)

From the second line, we can identify:
- Direction ratios: \( (3, 2, 3) \)
- Point on the line: \( (2, 3, 2) \)

### Step 2: Use the Condition for Coplanarity
For the two lines to lie in the same plane, the scalar triple product of the direction ratios and the vector connecting the two points must be zero.

Let \( A(3, 2, 1) \) and \( B(2, 3, 2) \) be points on the two lines. The vector \( \overrightarrow{AB} \) can be calculated as:
\[ \overrightarrow{AB} = B - A = (2 - 3, 3 - 2, 2 - 1) = (-1, 1, 1) \]

### Step 3: Set Up the Scalar Triple Product
The scalar triple product of vectors \( \overrightarrow{AB} \), \( \vec{d_1} = (2, 3, \lambda) \), and \( \vec{d_2} = (3, 2, 3) \) must be zero:
\[ \overrightarrow{AB} \cdot (\vec{d_1} \times \vec{d_2}) = 0 \]

### Step 4: Calculate the Cross Product
First, calculate the cross product \( \vec{d_1} \times \vec{d_2} \):
\[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & \lambda \\ 3 & 2 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}(3 \cdot 3 - \lambda \cdot 2) - \hat{j}(2 \cdot 3 - \lambda \cdot 3) + \hat{k}(2 \cdot 2 - 3 \cdot 3) \]
\[ = \hat{i}(9 - 2\lambda) - \hat{j}(6 - 3\lambda) + \hat{k}(4 - 9) \]
\[ = (9 - 2\lambda, -6 + 3\lambda, -5) \]

### Step 5: Dot Product with \( \overrightarrow{AB} \)
Now, take the dot product with \( \overrightarrow{AB} = (-1, 1, 1) \):
\[ (-1)(9 - 2\lambda) + (1)(-6 + 3\lambda) + (1)(-5) = 0 \]
\[ -(9 - 2\lambda) - 6 + 3\lambda - 5 = 0 \]
\[ -9 + 2\lambda - 6 + 3\lambda - 5 = 0 \]
\[ 5\lambda - 20 = 0 \]
\[ \lambda = 4 \]

### Step 6: Find the Point of Intersection
Now substitute \( \lambda = 4 \) back into the first line's equation:
\[ \frac{x-3}{2} = \frac{y-2}{3} = \frac{z-1}{4} = k \]
This gives:
\[ x = 2k + 3, \quad y = 3k + 2, \quad z = 4k + 1 \]

### Step 7: Substitute into the Second Line
Now substitute these into the second line's equation:
\[ \frac{2k + 3 - 2}{3} = \frac{3k + 2 - 3}{2} = \frac{4k + 1 - 2}{3} \]
This simplifies to:
\[ \frac{2k + 1}{3} = \frac{3k - 1}{2} = \frac{4k - 1}{3} \]

### Step 8: Solve for \( k \)
Setting the first two equal:
\[ 2(2k + 1) = 3(3k - 1) \]
\[ 4k + 2 = 9k - 3 \]
\[ 5k = 5 \implies k = 1 \]

### Step 9: Find Coordinates
Substituting \( k = 1 \) back:
\[ x = 2(1) + 3 = 5, \quad y = 3(1) + 2 = 5, \quad z = 4(1) + 1 = 5 \]
Thus, the point of intersection is \( (5, 5, 5) \).

### Step 10: Check Which Plane Contains the Point
Now we need to check which plane contains the point \( (5, 5, 5) \) from the options given.