14. Given the sets $A=\left\{(x, y) \left\lvert\, \frac{y-3}{x-2}=a+1\right.\right\}, B=\left\{(x, y) \mid\left(a^{2}-1\right) x+\right.$ $(a-1) y=15\}$. If $A \cap B=\varnothing$, find the value of the real number $a$.
14. $A=\{(x, y) \mid y-3=(a+1)(x-2), x \neq 2\}$, the set $A$ can be regarded as the line $l_{1}: y-3=(a+1)(x-2)$, excluding the point $(2,3)$. When $a=1$, $B=\varnothing$, then $A \cap B=\varnothing$ satisfies the condition; When $a \neq 1$, the set $B$ can be regarded as the line $l_{2}: y=-(a+1)x+\frac{15}{a-1}$. Then when $l_{1} / / l_{2}$, $A \cap B=\varnothing$, i.e., $a+1=-(a+1)$, and $3-2(a+1) \neq \frac{15}{a-1}$, we get $a=-1$; When $l_{2}$ passes through the point $(2,3)$, $A \cap B=\varnothing$, i.e., $2(a^2-1)+3(a-1)=15$, we get $a=\frac{5}{2}$ or -4. In summary, $a=1,-1, \frac{5}{2},-4$.