If the line `(x-1)/(-3) = (y-2)/(2k) = (z-3)/( 2) and (x-1)/( 3k) = (y-5)/(1) = (6-z)/(5)` are mutually perpendicular, then k is equal to

Options

A. `(10)/(7)`
B. `(-10)/(7)`
C. `(-10)`
D. `7`

Answer

To solve the problem, we need to determine the value of \( k \) such that the two given lines are mutually perpendicular.

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

The first line \( L_1 \) is given by:

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

From this, we can extract the direction ratios of line \( L_1 \):
- Direction ratios for \( L_1 \): \( (-3, 2k, 2) \)

The second line \( L_2 \) is given by:

\[ \frac{x-1}{3k} = \frac{y-5}{1} = \frac{6-z}{5} \]

From this, we can extract the direction ratios of line \( L_2 \):
- Direction ratios for \( L_2 \): \( (3k, 1, -5) \)

### Step 2: Use the condition for perpendicularity

For two lines to be mutually perpendicular, the dot product of their direction ratios must equal zero. Thus, we have:

\[ (-3)(3k) + (2k)(1) + (2)(-5) = 0 \]

### Step 3: Simplify the equation

Expanding the dot product gives:

\[ -9k + 2k - 10 = 0 \]

Combining like terms:

\[ -7k - 10 = 0 \]

### Step 4: Solve for \( k \)

Now, we can isolate \( k \):

\[ -7k = 10 \]

Dividing both sides by -7:

\[ k = -\frac{10}{7} \]

### Conclusion

Thus, the value of \( k \) is:

\[ \boxed{-\frac{10}{7}} \]