Answer: A
The maximum length of the rod that can be placed inside the room = the diagonal of the room \( = \sqrt{5^2 + 4^2 + 3^2} \) meters \( = \sqrt{25 + 16 + 9} \) meters \( = \sqrt{50} \) meters \( = 5\sqrt{2} \) meters
The maximum length of the rod that can be placed inside the room = the diagonal of the room \( = \sqrt{5^2 + 4^2 + 3^2} \) meters \( = \sqrt{25 + 16 + 9} \) meters \( = \sqrt{50} \) meters \( = 5\sqrt{2} \) meters