The value of k for which the lines
`(x+1)/(-3)=(y+5)/(2k) =(z-4)/(2)` and `(x-3)/(3k)=(y-2)/(1)=(z+1)/(7)` are perpendicular is

Options

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

Answer

To find the value of \( k \) for which the lines

\[ \frac{x+1}{-3} = \frac{y+5}{2k} = \frac{z-4}{2} \]

and

\[ \frac{x-3}{3k} = \frac{y-2}{1} = \frac{z+1}{7} \]

are perpendicular, we will follow these steps:

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

For the first line, we can express it in terms of a parameter \( t \):

\[ x = -3t - 1, \quad y = 2kt - 5, \quad z = 2t + 4 \]

From this, we can extract the direction vector \( \mathbf{d_1} \):

\[ \mathbf{d_1} = \langle -3, 2k, 2 \rangle \]

For the second line, we can express it in terms of a parameter \( s \):

\[ x = 3ks + 3, \quad y = s + 2, \quad z = 7s - 1 \]

From this, we can extract the direction vector \( \mathbf{d_2} \):

\[ \mathbf{d_2} = \langle 3k, 1, 7 \rangle \]

### Step 2: Use the dot product to determine perpendicularity

Two lines are perpendicular if the dot product of their direction vectors is zero:

\[ \mathbf{d_1} \cdot \mathbf{d_2} = 0 \]

Calculating the dot product:

\[ \mathbf{d_1} \cdot \mathbf{d_2} = (-3)(3k) + (2k)(1) + (2)(7) \]

This simplifies to:

\[ -9k + 2k + 14 = 0 \]

### Step 3: Solve the equation

Combining like terms gives:

\[ -7k + 14 = 0 \]

Now, isolate \( k \):

\[ -7k = -14 \]

Dividing both sides by -7:

\[ k = 2 \]

### Conclusion

The value of \( k \) for which the lines are perpendicular is

\[ \boxed{2} \]

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