12. (10 points) There are four colors of light bulbs (with an ample supply of each color), and one light bulb needs to be installed at each vertex of the triangular prism $A B C-A_{1} B_{1} C_{1}$. The requirement is that the light bulbs at the two endpoints of the same edge must be of different colors, and each color of light bulb must be used at least once. How many installation methods are there?
【Analysis】According to the problem, it is divided into 3 steps. In the first step, choose the color of the light bulbs for points $A$, $B$, and $C$, and the number of cases can be obtained using the permutation formula. In the second step, choose one of $A_{1}$, $B_{1}$, and $C_{1}$ to install the fourth color of the light bulb. In the third step, choose colors for the remaining two lights, and the number of cases can be obtained through classification discussion. Then, by the principle of step-by-step counting, the answer can be calculated.

【Solution】According to the problem, each color of the light bulb is used at least once, meaning four different colors of lights are used for installation. It is divided into 3 steps,

Step 1: Choose three colors of light bulbs for points $A$, $B$, and $C$, with a total of $A_{4}^{3}$ ways;
Step 2: Choose one of $A_{1}$, $B_{1}$, and $C_{1}$ to install the fourth color of the light bulb, with 3 possible choices;
Step 3: Choose colors for the remaining two lights,
Assuming the remaining are $B_{1}$ and $C_{1}$, if $B_{1}$ is the same color as $A$,
then $C_{1}$ can only choose the color of point $B$; if $B_{1}$ is the same color as $C$,
then $C_{1}$ has two colors to choose from, which are the colors of points $A$ and $B$.
Therefore, there are 3 ways to choose the light bulbs for $B_{1}$ and $C_{1}$,
which means there are 3 ways for the remaining two lights,
Thus, there are a total of $A_{4}^{3} \times 3 \times 3 = 216$ ways.
Answer: There are 216 ways to install the lights.