In sub-parts (i) and (ii) choose the correct option and in sub-parts (iii) to (v), answer the questions as instructed.
Find k so that the lines `(1-x)/(3)=(y-2)/(2k)=(z-3)/(2)` and `(x-1)/(3k)=(y-5)/(1)=(6-z)/(5)`are at right angles .

Answer

To find the value of \( k \) such that the lines

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

and

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

are at right angles, we can follow these steps:

### Step 1: Identify Direction Ratios
For the first line \( L_1 \):
\[ \frac{1-x}{3} = \frac{y-2}{2k} = \frac{z-3}{2} \]
The direction ratios can be derived from the coefficients:
- For \( x \): \( -3 \)
- For \( y \): \( 2k \)
- For \( z \): \( 2 \)

Thus, the direction ratios of \( L_1 \) are \( (-3, 2k, 2) \).

For the second line \( L_2 \):
\[ \frac{x-1}{3k} = \frac{y-5}{1} = \frac{6-z}{5} \]
The direction ratios are:
- For \( x \): \( 3k \)
- For \( y \): \( 1 \)
- For \( z \): \( -5 \)

Thus, the direction ratios of \( L_2 \) are \( (3k, 1, -5) \).

### Step 2: Use the Condition for Perpendicularity
Two lines are perpendicular if the dot product of their direction ratios is zero. Therefore, we need to calculate:

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

Calculating the dot product:
\[ -3(3k) + 2k(1) + 2(-5) = 0 \]
This simplifies to:
\[ -9k + 2k - 10 = 0 \]

### Step 3: Solve for \( k \)
Combining like terms:
\[ -7k - 10 = 0 \]
Adding \( 10 \) to both sides:
\[ -7k = 10 \]
Dividing by \( -7 \):
\[ k = -\frac{10}{7} \]

### Final Answer
Thus, the value of \( k \) such that the lines are at right angles is:
\[ \boxed{-\frac{10}{7}} \]