Fiind m, if the lines `(1-x)/(3)= ( 7y-14)/(2m ) = (z-3) /(2) and ( 7-7x)/(3m) = (y-5)/(1) = (6-z)/(5)` are at right angles.

Answer

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

### Step 1: Write the equations in standard form
The given equations are:
\[ \frac{1-x}{3} = \frac{7y-14}{2m} = \frac{z-3}{2} \]
and
\[ \frac{7-7x}{3m} = \frac{y-5}{1} = \frac{6-z}{5} \]

We can rewrite these equations in a standard parametric form.

For the first line:
\[ \frac{1-x}{3} = t \implies x = 1 - 3t \]
\[ \frac{7y-14}{2m} = t \implies 7y - 14 = 2mt \implies y = \frac{2mt + 14}{7} \]
\[ \frac{z-3}{2} = t \implies z = 2t + 3 \]

Thus, the direction ratios \( \text{dR}_1 \) for the first line are:
\[ \text{dR}_1 = (-3, \frac{2m}{7}, 2) \]

For the second line:
\[ \frac{7-7x}{3m} = s \implies 7 - 7x = 3ms \implies x = 1 - \frac{3ms}{7} \]
\[ \frac{y-5}{1} = s \implies y = s + 5 \]
\[ \frac{6-z}{5} = s \implies 6 - z = 5s \implies z = 6 - 5s \]

Thus, the direction ratios \( \text{dR}_2 \) for the second line are:
\[ \text{dR}_2 = (-3m, 1, -5) \]

### Step 2: Use the condition for perpendicular lines
For the lines to be perpendicular, the dot product of their direction ratios must equal zero:
\[ \text{dR}_1 \cdot \text{dR}_2 = 0 \]

Calculating the dot product:
\[ (-3) \cdot (-3m) + \left(\frac{2m}{7}\right) \cdot 1 + 2 \cdot (-5) = 0 \]
This simplifies to:
\[ 9m + \frac{2m}{7} - 10 = 0 \]

### Step 3: Solve for \( m \)
To eliminate the fraction, multiply the entire equation by 7:
\[ 63m + 2m - 70 = 0 \]
Combine like terms:
\[ 65m - 70 = 0 \]
Thus,
\[ 65m = 70 \implies m = \frac{70}{65} = \frac{14}{13} \]

### Final Answer
The value of \( m \) is:
\[ m = \frac{14}{13} \]