3. As shown in Figure 2, given the line $y=\frac{1}{2} x$ intersects the parabola $y=-\frac{1}{4} x^{2}+6$ at points $A$ and $B$, point $P$ moves on the parabola above line $A B$. When the area of $\triangle P A B$ is maximized, the coordinates of point $P$ are $\qquad$
3. $\left(-1, \frac{23}{4}\right)$.

As shown in Figure 7, construct $P M \perp x$-axis intersecting line $A B$ at point $M$.
Let $P\left(a,-\frac{1}{4} a^{2}+6\right)$.
Then $M\left(a, \frac{1}{2} a\right)$.
Thus, $P M=-\frac{1}{4} a^{2}-\frac{1}{2} a+6$.
By solving $y=\frac{1}{2} x$ and $y=-\frac{1}{4} x^{2}+6$ simultaneously, we find that the x-coordinates of points
$A$ and $B$ are $-6$ and $4$, respectively.
And $S_{\triangle P A B}=S_{\triangle P A M}+S_{\triangle P B M}$
$=\frac{1}{2}(6+4) P M=-\frac{5}{4}(a+1)^{2}+\frac{125}{4}$.
Therefore, when $a=-1$, $y_{\text {max }}=\frac{125}{4}$.
At this point, $P\left(-1, \frac{23}{4}\right)$.