<s>[INST] $6$ people, $3$ rooms, $1$ opening door $6$ people spread in $3$ distinguishable rooms, every room needs one person who opens the door.
There are ${6 \choose 3}\cdot 3 \cdot 2$ options to choose the three door opener persons and let them open one certain room, so that every room is opened by one person.
Further, there are then $3$ options to have all left three guys in one of the rooms, $3\cdot 2$ options to have exactly one other guy in every room, and ${3 choose 2}\cdot 3 \cdot 2$ options to have one guy in one room, two in the other and none in the last room.
So in total there are ${6 \choose 3}\cdot 3 \cdot 2 \cdot 3 \cdot 3 \cdot 2 {3 \choose 2} \cdot 3 \cdot 2$ options to have all rooms opened by exactly one person and spread the others in these rooms.
If only there are 3 people who can open the doors (only they have keys for all the rooms). There are only $3 \cdot 2 \cdot 3 \cdot 3 \cdot 2 {3 \choose 2} \cdot 3 \cdot 2$ options left, right?
 [/INST] Here's an easy way to look at the problem:
First, we make sure that we have a door opener in each room.  There are $6$ choices for the first room, $5$ for the second, and $4$ for the third.  Thus, we may select door openers in $6\times5\times4=120$ ways.
Now, each of the three remaining people can be put in any of the three rooms.  So, we have $3\times3\times3=27$ ways of distributing the remaining people after the door openers are chosen.
Thus, the total number of possibilities is $120\times27=3240$

As far as I can tell, you've followed the same steps, but I think this is a bit less redundant.  For example, instead of writing that there are $\binom{6}{3}\times3\times2$ ways of selecting door openers, you could have noticed that
$$
\begin{align}
\binom{6}{3}\times3\times2 &= \frac{6!}{3!3!}\times3!\\
&=\frac{6!}{3!}\\
&=\frac{6\times5\times4\times3\times2}{3\times2}\\
&=6\times5\times4
\end{align}
$$
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