Solution:

In counting the number of sets of $3$ cards, we first want to choose which of our two attributes will be good and which of our two attributes will not be good. There are $\binom{4}{2} = 6$ such choices.

Now consider the two attributes which are not good, attribute $X$ and attribute $Y$. Since these are not good, some value should appear exactly twice. Suppose the value $a$ appears twice and $b$ appears once for attribute $X$ and that the value $c$ appears twice and $d$ appears once for attribute $Y$. There are three choices for $a$ and then two choices for $b$; similarly, there are three choices for $c$ and then two choices for $d$. This gives $3 \cdot 2 \cdot 3 \cdot 2 = 36$ choices of $a, b, c$, and $d$.

There are two cases to consider. The first is that there are two cards which both have $a$ and $c$, while the other card has both $b$ and $d$. The second case is that only one card has both $a$ and $c$, while one card has $a$ and $d$ and the other has $b$ and $c$.

Case 1:

The three cards need to be distinct. Card 3 is necessarily distinct from Card 1 and Card 2, but we need to ensure that Card 1 and Card 2 are distinct from each other. There are $9$ choices for the two good attributes of Card 1, and then $8$ choices for the two good attributes of Card 2. But we also want to divide by $2$ since we do not care about the order of Card 1 and Card 2. So there are $\frac{9 \cdot 8}{2} = 36$ choices for the good attributes on Card 1 and Card 2. Then, the values of the good attributes of Card 1 and Card 2 uniquely determine the values of the good attributes of Card 3.

Case 2:

Card 1, Card 2, and Card 3 will all be distinct no matter what the values of the good attributes are, because the values of attributes $X$ and $Y$ are unique to each card. So there are $9$ possibilities for the values of the good attributes on Card 1, and then there are $9$ more possibilities for the values of the good attributes on Card 2. We do not have to divide by $2$ this time, since Card 1 and Card 2 have distinct values in $X$ and $Y$. So there are $9^2 = 81$ possibilities here.

So our final answer is $6 \cdot 6^2 \cdot (36 + 81) = 25272$.