Step 1: The rates of the pipes are as follows: Pipe A (inlet): \( \frac{1}{3} \) of the cistern per hour. Pipe B (inlet): \( \frac{1}{4} \) of the cistern per hour. Pipe C (outlet): \( \frac{1}{1} = 1 \) of the cistern per hour.
Step 2: Between 5 a.m. and 6 a.m. (Only Pipe A is open): In 1 hour, Pipe A fills: \[ \frac{1}{3} \, {of the cistern}. \]
Step 3: Between 6 a.m. and 7 a.m. (Pipes A and B are open): Combined rate of Pipes A and B: \[ \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \, {of the cistern per hour}. \] In 1 hour, Pipes A and B together fill: \[ \frac{7}{12} \, {of the cistern}. \]
Step 4: At 7 a.m., Pipe C is also opened (Pipes A, B, and C are all open): Combined rate of Pipes A, B, and C: \[ \frac{1}{3} + \frac{1}{4} - 1 = \frac{7}{12} - 1 = \frac{-5}{12} \, {of the cistern per hour}. \] This means the cistern is being emptied at a rate of \( \frac{5}{12} \) of the cistern per hour.
Step 5: Total filled by 7 a.m.: From 5 a.m. to 6 a.m. (1 hour by Pipe A): \[ \frac{1}{3} \, {of the cistern}. \] From 6 a.m. to 7 a.m. (1 hour by Pipes A and B): \[ \frac{7}{12} \, {of the cistern}. \] Total filled by 7 a.m.: \[ \frac{1}{3} + \frac{7}{12} = \frac{4}{12} + \frac{7}{12} = \frac{11}{12} \, {of the cistern}. \]
Step 6: Time to empty the cistern after 7 a.m.: At 7 a.m., the cistern is \( \frac{11}{12} \) full, and it is being emptied at a rate of \( \frac{5}{12} \) per hour. Time required to empty \( \frac{11}{12} \) of the cistern: \[ {Time} = \frac{{Volume to Empty}}{{Rate of Emptying}} = \frac{\frac{11}{12}}{\frac{5}{12}} = \frac{11}{5} \, {hours} = 2.2 \, {hours}. \]
Step 7: Final time: 2.2 hours after 7 a.m. is: \[ 7:00 \, {a.m.} + 2 \, {hours and} \, 12 \, {minutes} = 9:12 \, {a.m.}. \]
Final Answer: The cistern will be empty at \( \mathbf{9:12 \, {a.m.}} \).
The system of simultaneous linear equations :
\[ \begin{array}{rcl} x - 2y + 3z &=& 4 \\ 2x + 3y + z &=& 6 \\ 3x + y - 2z &=& 7 \end{array} \]