11. Let the set $A=\left\{(x, y) \left\lvert\, \frac{y-3}{x-1}=2\right., x, y \in \mathbf{R}\right\}, B=\{(x, y) \mid 4 x+a y=16, x, y \in \mathbf{R}\}$, if $A \cap B=\varnothing$, then the value of the real number $a$ is $\qquad$.
11. Solve $A=\{(x, y) \mid y=2 x+1, x \neq 1\}$, which means the set $A$ consists of points on the line $y=2 x+1$ except for the point $(1,3)$. Therefore, when $-\frac{4}{a}=2$, i.e., $a=-2$, the set $B$ represents a line parallel to $y=2 x+1$, and at this time, $A \cap B=\varnothing$; Also, $(1,3) \in A$, so when $(1,3) \in B$, $A \cap B=\varnothing$, i.e., $1 \times 4+3 a-16=0$, thus $a=4$. In summary, the values of $a$ are -2 or 4.