If the lines `6x-2=3y+1=2z-2` and `(x-2)/(lamda)=(2y-5)/(-3),z=-2` are perpendicular then `lamda=`

Options

A. 3
B. 2
C. -3
D. 1

Explanation

To find the value of \(\lambda\) such that the lines \(6x-2=3y+1=2z-2\) and \(\frac{x-2}{\lambda}=\frac{2y-5}{-3}, z=-2\) are perpendicular, we can follow these steps:

### Step 1: Convert the first line into standard form
The first line is given as:
\[ 6x - 2 = 3y + 1 = 2z - 2 \]
We can express this in the standard form:
\[ \frac{x - \frac{1}{3}}{\frac{1}{6}} = \frac{y + \frac{1}{3}}{\frac{1}{3}} = \frac{z - 1}{\frac{1}{2}} \]
From this, we can identify the direction ratios of the first line as:
\[ (1, 2, 3) \]

### Step 2: Convert the second line into standard form
The second line is given as:
\[ \frac{x-2}{\lambda} = \frac{2y-5}{-3}, \quad z = -2 \]
We can express this in standard form:
\[ \frac{x - 2}{\lambda} = \frac{y - \frac{5}{2}}{-\frac{3}{2}} = \frac{z + 2}{0} \]
From this, we can identify the direction ratios of the second line as:
\[ (\lambda, -\frac{3}{2}, 0) \]

### Step 3: Use the condition for perpendicularity
For the two lines to be perpendicular, the dot product of their direction ratios must equal zero:
\[ (1, 2, 3) \cdot (\lambda, -\frac{3}{2}, 0) = 0 \]
Calculating the dot product:
\[ 1 \cdot \lambda + 2 \cdot \left(-\frac{3}{2}\right) + 3 \cdot 0 = 0 \]
This simplifies to:
\[ \lambda - 3 = 0 \]

### Step 4: Solve for \(\lambda\)
From the equation \(\lambda - 3 = 0\), we find:
\[ \lambda = 3 \]

### Conclusion
Thus, the value of \(\lambda\) is:
\[ \lambda = 3 \]