12. In a regular hexagon divided into six regions for planting ornamental plants (as shown in the figure), it is required that the same type of plant be planted in the same region, and different plants be planted in adjacent regions. Given 4 different types of plants to choose from, there are $\qquad$ planting schemes.

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12. In a regular hexagon divided into six regions for planting ornamental plants (as shown in the figure), it is required that the same type of plant be planted in the same region, and different plants be planted in adjacent regions. Given 4 different types of plants to choose from, there are $\qquad$ planting schemes.
732
12.【Analysis and Solution】For the convenience of narration, we label the six areas with letters $A, B, C, D, E, F$ respectively. The number of plant species planted in the areas $\mathrm{A}, \mathrm{C}, \mathrm{E}$, which are three apart, is divided into the following three categories.
(1) If $A, C, E$ are planted with the same species, there are 4 ways to do so. After planting $A, C, E$, $B, D, F$ can each choose one species from the remaining three species (repetition allowed), each having 3 methods. At this point, there are a total of $4 \times 3 \times 3 \times 3=108$ methods.
(2) If $A, C, E$ are planted with two species, there are $A_{4}^{2}$ ways to do so. After planting $A, C, E$, if $A, C$ are the same, then $B$ has 3 methods, and $D, F$ each have 2 methods; if $C, E$ or $E, A$ are the same, it is the same (just in a different order). At this point, there are a total of $A_{4}^{3} \times 3$ $(3 \times 2 \times 2)=432$ methods.
(3) If $A, C, E$ are planted with three species, there are $A_{4}^{3}$ ways to do so. At this time, $B, D, F$ each have 2 methods. At this point, there are a total of $A_{4}^{3} \times 2$ $\times 2 \times 2=192$ methods.
According to the principle of addition, there are a total of $N=108+432+192=732$ planting schemes.