(i) Find the value of 'p' so that the lines :
`l_(1) : (1 - x)/(3) = (7y -14)/(2p) = (z - 3)/(2) and l_(2) : (7 -7x)/(3p) = (y - 5)/(1) = (6 - z)/(5)` are at right angles.
Also, find the equations of the line passing through (3,2, -4) and parallel to line `l_(1)`.
(ii) Find 'k' so that the lines :
`(x - 3)/(2) = (y + 1)/(3 ) = (z - 2)/(2k) and (x + 2)/(1) = (4 -y)/(k) = (z + 5)/(1)`
are perpendicular to each other.

Answer

To solve the given problems step by step, we will break them down into manageable parts.

### Part (i)

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

For line \( l_1 \):
\[ \frac{1 - x}{3} = \frac{7y - 14}{2p} = \frac{z - 3}{2} \]
The direction ratios can be extracted as follows:
- From \( \frac{1 - x}{3} \), we have \( a_1 = -3 \).
- From \( \frac{7y - 14}{2p} \), we have \( b_1 = \frac{7}{2p} \).
- From \( \frac{z - 3}{2} \), we have \( c_1 = 2 \).

Thus, the direction ratios of line \( l_1 \) are \( (-3, \frac{7}{2p}, 2) \).

For line \( l_2 \):
\[ \frac{7 - 7x}{3p} = \frac{y - 5}{1} = \frac{6 - z}{5} \]
The direction ratios can be extracted as follows:
- From \( \frac{7 - 7x}{3p} \), we have \( a_2 = -\frac{7}{3p} \).
- From \( \frac{y - 5}{1} \), we have \( b_2 = 1 \).
- From \( \frac{6 - z}{5} \), we have \( c_2 = -5 \).

Thus, the direction ratios of line \( l_2 \) are \( (-\frac{7}{3p}, 1, -5) \).

**Step 2: Set up the condition for perpendicularity**

Two lines are perpendicular if the dot product of their direction ratios is zero:
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \]
Substituting the values:
\[ (-3) \left(-\frac{7}{3p}\right) + \left(\frac{7}{2p}\right)(1) + (2)(-5) = 0 \]

**Step 3: Simplify the equation**

Calculating each term:
\[ 7/p + \frac{7}{2p} - 10 = 0 \]
Multiplying through by \( 2p \) to eliminate the denominators:
\[ 14 + 7 - 20p = 0 \]
This simplifies to:
\[ 21 = 20p \implies p = \frac{21}{20} \]

**Step 4: Find the equation of the line parallel to \( l_1 \) through the point (3, 2, -4)**

The equation of a line in the form \( \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1} \) can be written as:
\[ \frac{x - 3}{-3} = \frac{y - 2}{\frac{7}{2p}} = \frac{z + 4}{2} \]
Substituting \( p = \frac{21}{20} \):
\[ \frac{x - 3}{-3} = \frac{y - 2}{\frac{7}{2 \cdot \frac{21}{20}}} = \frac{z + 4}{2} \]
This gives us the final equation of the line.

### Part (ii)

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

For line \( l_1 \):
\[ \frac{x - 3}{2} = \frac{y + 1}{3} = \frac{z - 2}{2k} \]
The direction ratios are \( (2, 3, 2k) \).

For line \( l_2 \):
\[ \frac{x + 2}{1} = \frac{4 - y}{k} = \frac{z + 5}{1} \]
The direction ratios are \( (1, -k, 1) \).

**Step 2: Set up the condition for perpendicularity**

Using the same dot product condition:
\[ 2 \cdot 1 + 3 \cdot (-k) + 2k \cdot 1 = 0 \]
This simplifies to:
\[ 2 - 3k + 2k = 0 \implies 2 - k = 0 \implies k = 2 \]

### Summary of Results
- For part (i), \( p = \frac{21}{20} \) and the equation of the line through (3, 2, -4) parallel to \( l_1 \) is derived.
- For part (ii), \( k = 2 \).