Given, two taps L and M can fill a tank in 15 and 12 hours.

In 1 hr,

Tap L fills 1/15^th of the tank and tap M fills 1/12^th of the tank

Also, tap N can empty the tank in 4 hour.

In 1 hr, tap N empties 1/4^th of the tank.

Let the number of hours taken to empty the tank be h.

Given, taps L, M and N are opened at 9 am, 11 am and 12 pm, respectively.

Tank will be emptied at the end of ‘h’ hours after 9 am.

∴ Number of hours for which tank M is opened = h – 2

Number of hours for which tank N is opened = h – 3

$(\begin{array}{l} \therefore \frac{h}{{15}} + \frac{{h - 2}}{{12}} - \frac{{h - 3}}{4} = 0\\ \Rightarrow \;\frac{{4h + 5h - 10 - 15h + 45}}{{60}} = 0 \end{array})$

⇒ 6h = 35

⇒ h = 35/6 hr = 5 hr 50 min

Thus from 9 am it takes 5 hr 50 min to empty the tank.