We have $\text{LILAC}$.
Lets number the positions as $1,2,3,4,5$. Now the two $L$s cannot be placed at position $1$ and $3$ but they can be positioned at $2,4,5$ in $^3C_2= 3$ ways. (since the $Ls$ are indistinguishable)
Now one of $2,4,5$ is vacant. Without loss of generality lets say $2$ is vacant.
Now, if $2$ is vacant we can't palce $I$ there but we can place any of $A$, or $C$, so we have $2$ choices for the position which is left after filling the two $Ls$. Now all of $2,4,5$ are filled.
For the remaining two places $1,3$ we have two characters left and none of them is $L$ so we can place them in $2! = 2 $ ways.
Multiply them all = $^3C_2 * 2 * 2! = 3 * 2 * 2 = 12 \ (ans)$.