Find the value of p so that the lines `(1-x)/(3)= (7y-14)/(2p)= (z-3)/(2) and (7-7x)/(3p)= (y-5)/(1)= (6-z)/(5)` are at right angles

Answer

To find the value of \( p \) such that the given lines are at right angles, we will follow these steps:

### Step 1: Write the equations of the lines in parametric form
The first line is given by:
\[ \frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2} \]
Let \( t \) be the parameter. Then we can express the coordinates as:
\[ x = 1 - 3t, \quad y = 2 + \frac{2pt}{7}, \quad z = 3 + 2t \]

### Step 2: Identify direction ratios of the first line
From the parametric equations, we can identify the direction ratios \( (a_1, b_1, c_1) \) of the first line:
\[ a_1 = -3, \quad b_1 = \frac{2p}{7}, \quad c_1 = 2 \]

### Step 3: Write the second line in parametric form
The second line is given by:
\[ \frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5} \]
Let \( s \) be the parameter. Then we can express the coordinates as:
\[ x = 1 - \frac{3ps}{7}, \quad y = 5 + s, \quad z = 6 - 5s \]

### Step 4: Identify direction ratios of the second line
From the parametric equations, we can identify the direction ratios \( (a_2, b_2, c_2) \) of the second line:
\[ a_2 = -\frac{3p}{7}, \quad b_2 = 1, \quad c_2 = -5 \]

### Step 5: Use the condition for perpendicularity
The lines are perpendicular if:
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \]
Substituting the values we found:
\[ (-3) \left(-\frac{3p}{7}\right) + \left(\frac{2p}{7}\right)(1) + (2)(-5) = 0 \]

### Step 6: Simplify the equation
This simplifies to:
\[ \frac{9p}{7} + \frac{2p}{7} - 10 = 0 \]
Combining the terms gives:
\[ \frac{11p}{7} - 10 = 0 \]

### Step 7: Solve for \( p \)
Rearranging the equation:
\[ \frac{11p}{7} = 10 \]
Multiplying both sides by 7:
\[ 11p = 70 \]
Dividing by 11:
\[ p = \frac{70}{11} \]

### Final Answer
Thus, the value of \( p \) is:
\[ \boxed{\frac{70}{11}} \]