Let $A,B,C$ be the centers of the circles of radii $2,3,4$. Then $AB=2+3=5$, $AC=2+4=6$, $BC=3+4=7$.
Place $A=(0,0)$ and $B=(5,0)$. By the law of cosines, $\cos A=\dfrac{25+36-49}{2\cdot5\cdot6}=\dfrac15$, so $C=\left(6\cdot\dfrac15,\,6\cdot\dfrac{2\sqrt6}5\right)=\left(\dfrac65,\dfrac{12\sqrt6}5\right)$.
Each tangency point lies on the segment between the two centers, at distance equal to the radius from the center: $P$ on $AB$ at distance $2$ from $A$, $Q$ on $AC$ at distance $2$ from $A$, $R$ on $BC$ at distance $3$ from $B$. So $P=(2,0)$, $Q=\dfrac26C=\left(\dfrac25,\dfrac{4\sqrt6}5\right)$, $R=B+\dfrac37(C-B)=\left(\dfrac{118}{35},\dfrac{36\sqrt6}{35}\right)$.
With $Q-P=\left(-\dfrac85,\dfrac{4\sqrt6}5\right)$ and $R-P=\left(\dfrac{48}{35},\dfrac{36\sqrt6}{35}\right)$, the area is $$\frac12\left|\left(-\frac85\right)\frac{36\sqrt6}{35}-\frac{4\sqrt6}5\cdot\frac{48}{35}\right|=\frac12\cdot\frac{480\sqrt6}{175}=\boxed{\frac{48\sqrt6}{35}}\approx3.36.$$
