According to the problem, $BC=6$, with the midpoint of $BC$ as the origin and the line where $BC$ lies as the $x$-axis, we establish a rectangular coordinate system. Thus, $B(-3,0), C(3,0)$. Let $A(x, y)$, then
$\overrightarrow{A B} \cdot \overrightarrow{A C}=(x+3)(x-3)+y^{2}=x^{2}+y^{2}-9=7 \Rightarrow x^{2}+y^{2}=16$,
which means point $A$ lies on a circle centered at the origin with a radius of 4.
Therefore, the maximum area of $\triangle ABC$ is $\frac{1}{2} \times 6 \times 4=12$.
5. In $\triangle A B C$, if $\overrightarrow{A B} \cdot \overrightarrow{A C}=7,|\overrightarrow{A B}-\overrightarrow{A C}|=6$, then the maximum value of the area of $\triangle A B C$ is