3. If the orthocenter of $\triangle O A B$ is exactly the focus of the parabola $y^{2}=4 x$, where $O$ is the origin, and points $A$ and $B$ are on the parabola, then the area $S$ of $\triangle O A B$ is $\qquad$ .
3. $10 \sqrt{5}$.

Notice that, the focus of the parabola is $F(1,0)$. Since $F$ is the orthocenter of $\triangle O A B$, we have $O F \perp A B$.
Therefore, we can set $A\left(a^{2}, 2 a\right), B\left(a^{2},-2 a\right)(a>0)$.
Thus, $l_{O A}: a y=2 x \Rightarrow k_{O A}=\frac{2}{a}$.
Also, $k_{B F}=\frac{-2 a}{a^{2}-1}$, and from $k_{B F} k_{O A}=-1$, we get $a=\sqrt{5}$.
Therefore, $A B=4 \sqrt{5}, h=a^{2}=5$.
Hence, $S=10 \sqrt{5}$.