Solution:

6015
Let $a$ and $d$ be the first term and the common difference of the given arithmetic sequence. Since there are distinct terms of the sequence, it follows that $d \neq 0$. Since the third, fifth and eleventh terms form a geometric sequence, we have
$$\frac{a+4d}{a+2d} = \frac{a+10d}{a+4d} \quad \text{or} \quad a+d=0$$
Since the fourth term of the sequence is $6$, we also have
$$a+3d=6$$
Solving the system of equations involving $(1\star)$ and $(2\star)$, we get $a=-3$ and $d=3$. Thus, the $2007$th term is $-3+2006(3)=6015$.