Let us name the pipes as A, B and C
$T_A = 12$ hours
$T_B = 18$ hours
$T_C = 36$ hours

We know Rate, $R = \frac{1}{T}$
Therefore,
$R_A = \frac{1}{12}$
$R_B = \frac{1}{18}$
$R_C = \frac{1}{36}$

contractor opens the first pipe and after 2 hours, the second pipe. At the end of 6 hours, he realized that the third pipe is open too
i.e. A worked for 6 hours, B worked for 4 hours and C worked for 6 hours too

So, their combined work = $\frac{6}{12} + \frac{4}{18} - \frac{6}{36} = \frac{5}{9}$

(NOTE: -ve sign indicates that pipe C is doing opposite work)

Since, these pipes have filled $\frac{5}{9}^{th}$ of the tank in 6 hours, the remaining work is $1 - \frac{5}{9} = \frac{4}{9}$

i.e. $\frac{4}{9}^{th}$ of the work is to be completed by A and B together

$R_{A+B} = \frac{1}{12} + \frac{1}{18} = \frac{5}{36}$

$Rate = \frac{Work}{Time} = (\frac{4}{9})(\frac{36}{5}) = \frac{16}{5}$ hours

TOTAL time required to fill the tank = 6 hours + $\frac{16}{5}$ hours = $\frac{46}{5}$ hours = $\frac{46}{5}(60)$ minutes = $552$ minutes = $9$ hours $12$ minutes