Question 189: In a unit square, if two of its four sides are colored with color $i$, and the other two sides are colored with two different colors other than $i$, then the unit square is called “dominated by color $i$”. As shown in the figure, a $1 \times 3$ grid table contains a total of 10 unit line segments. Now, these 10 line segments are to be colored, each with one of the three colors: red, yellow, or blue, such that there is one unit square dominated by red, one by yellow, and one by blue. The number of such coloring methods is $\qquad$ _.
Question 189, Solution: For the coloring methods that meet the requirements, the following conclusions are obviously true:
(1) When one side of a square is colored with a certain color $\mathrm{i}$, if the square is dominated by color $\mathrm{i}$, then the other three sides have exactly $3!=6$ coloring methods: if the square is dominated by color $j(j \neq i)$, then the other three sides have 3 coloring methods;
(2) When two sides of a square are colored with color $\mathrm{i}$, the square can only be dominated by color $\mathrm{i}$, and the other two sides have exactly 2 coloring methods;
(3) When two sides of a square are colored with two colors $\mathrm{i}$ and $\mathrm{j}$, if the square is dominated by color $\mathrm{i}$ or $\mathrm{j}$, then the other two sides have exactly two coloring methods; if the square is dominated by color $k(\neq i, j)$, then the other two sides have a unique coloring method.

By symmetry, it is only necessary to find the number of coloring methods $N$ where the left, middle, and right squares are dominated by red, yellow, and blue, respectively, then the result of this problem is $6 \mathrm{~N}$. A coloring method is classified as “$\mathrm{i}-\mathrm{j}$ type” if the common side between the left and middle squares is colored with color $\mathrm{i}$, and the common side between the middle and right squares is colored with color $\mathrm{j}$. Using the conclusions (1), (2), and (3) above, the number of coloring methods for each type is shown in the table below:
Therefore, $\mathrm{N}=36 \times 3+18 \times 3+9=171$, so the number of coloring methods that meet the conditions is $6 \times 171=1026$.