Option C: 5 km
To solve this problem, we need to determine the distance from the worker's house to the factory. Let's denote the distance as \( d \) kilometers.
     Step 1: Establish equations based on given conditions.
         ○ When the worker travels at 5 km/hr, he is 3 minutes late. This means he takes 3 minutes more than the required time.
         ○ When the worker travels at 6 km/hr, he is 7 minutes early. This means he takes 7 minutes less than the required time.
     Let's denote the required time to reach the factory on time as \( t \) hours.
     Step 2: Convert minutes to hours.
         ○ 3 minutes = \(\frac{3}{60}\) hours = \(\frac{1}{20}\) hours
         ○ 7 minutes = \(\frac{7}{60}\) hours
     Step 3: Set up equations for each scenario.
         ○ At 5 km/hr: The time taken is \( \frac{d}{5} \) hours. Since he is 3 minutes late, the equation is:
       \[
       \frac{d}{5} = t + \frac{1}{20}
       \]
         ○ At 6 km/hr: The time taken is \( \frac{d}{6} \) hours. Since he is 7 minutes early, the equation is:
       \[
       \frac{d}{6} = t - \frac{7}{60}
       \]
     Step 4: Solve the equations.
         ○ From the first equation:
       \[
       t = \frac{d}{5} - \frac{1}{20}
       \]
         ○ From the second equation:
       \[
       t = \frac{d}{6} + \frac{7}{60}
       \]
     Step 5: Equate the two expressions for \( t \).
     \[
     \frac{d}{5} - \frac{1}{20} = \frac{d}{6} + \frac{7}{60}
     \]
     Step 6: Clear the fractions by finding a common denominator.
     The common denominator for 5, 6, 20, and 60 is 60. Multiply every term by 60 to eliminate the fractions:
     \[
     60 \left(\frac{d}{5}\right) - 60 \left(\frac{1}{20}\right) = 60 \left(\frac{d}{6}\right) + 60 \left(\frac{7}{60}\right)
     \]
     Simplifying gives:
     \[
     12d - 3 = 10d + 7
     \]
     Step 7: Solve for \( d \).
     \[
     12d - 10d = 7 + 3
     \]
     \[
     2d = 10
     \]
     \[
     d = 5
     \]
     Therefore, the distance of the factory from his house is 5 km.