3. A line $l$ is drawn through the fixed point $P(2,1)$, intersecting the positive direction of the $x$-axis and the positive direction of the $y$-axis at points $A$ and $B$, respectively, such that the area of $\triangle A O B$ (where $O$ is the origin) is minimized. The equation of $l$ is ( ).
(A) $x+y-3=0$
(B) $x+3 y-5=0$
(C) $2 x+y-5=0$
(D) $x+2 y-4=0$
3. (D).

It is easy to get $S_{\triangle A n B}=\frac{1}{2} a b$, and we have $\frac{2}{a}+\frac{1}{b}=1$, which means $a b=a+$ $2 b \geqslant 2 \sqrt{a \cdot 2 b}=2 \sqrt{2} \sqrt{a b}$. Therefore, $\sqrt{a b} \geqslant 2 \sqrt{2}, a b \geqslant 8$. The equality holds if and only if $\left\{\begin{array}{l}a=2 b, \\ 2 b+a=a b,\end{array}\right.$ which is $\left\{\begin{array}{l}a=4, \\ b=2\end{array}\right.$, so the equation of $l$ is $\frac{x}{4}+\frac{y}{2}=1$, or $x+2 y-4=0$.