TE. ** Given the set $A=\left\{(x, y) \left\lvert\, \frac{y-3}{x-2}=\right.\right.$
$$
a+1\}, B=\left\{(x, y) \mid\left(a^{2}-1\right) x+(a-1) y=\right.
$$
15\}. Question: For what value of the real number $a$ is $A \cap B=\varnothing$?
When $a=1$, $B=\varnothing$, which meets the requirement. When $a \neq 1$, set $A$ represents the line $y=(a+1)x-2a+1$ (where $x \neq 2$), and $B$ represents the line $y=-(a+1)x+\frac{15}{a-1}$. Since $A \cap B=\varnothing$, it means these two lines are either parallel or intersect at a point $P$ where the x-coordinate of $P$ is $x=2$. The former requires $a+1=-(a+1)$ and $-2a+1+\frac{15}{a-1}$, while the latter requires $2(a+1)-2a+1=-2(a+1)+\frac{15}{a-1}$. Solving these respectively yields $a=-1$ or $a \in\left\{\frac{5}{2},-4\right\}$. Therefore, when $a \in\left\{-1,-4,1, \frac{5}{2}\right\}$, $A \cap B=\varnothing$.