(3) For the regular triangular pyramid $V-ABC$, the side edge length is 3, and the base edge length is 2. A section through the base edge $AB$ intersects the side edge $VC$ at point $D$. The minimum value of the area of the section $\triangle ABD$ is $\qquad$.
(3) $\frac{\sqrt{23}}{3}$ Hint: Let the midpoint of $A B$ be $E$, it is easy to know that $A B \perp$ plane $V E C$, and the projection of $V$ on the base plane $O$ lies on $C E$, so $D E \perp A B, V O=\sqrt{V C^{2}-O C^{2}}=$ $\sqrt{\frac{23}{3}}$. To minimize the area of $\triangle A B D$, $D E$ must be minimized, i.e., $D E$ is the common perpendicular of $A B$ and $V C$.
In $\triangle V C E$, by the equal area method, we find $D E=\frac{\sqrt{23}}{3}$. Therefore,
$$
S_{\triangle A B D}=\frac{1}{2} A B \cdot D E=\frac{\sqrt{23}}{3} .
$$