Let $r_{1}, r_{2}$, and $r_{3}$ be the radii of the three circles. Since the circles are tangent to each other in pairs, we have

$$
\left\{\begin{array}{l}
r_{1}+r_{2}=3 \\
r_{1}+r_{3}=4 \\
r_{2}+r_{3}=5
\end{array}\right.
$$

Substituting the values $r_{2}=3-r_{1}$ and $r_{3}=4-r_{1}$ into the third equation, we get $3-r_{1}+4-r_{1}=5$. Therefore, $r_{1}=1, r_{2}=2$, and $r_{3}=3$. Thus, the sum of the areas of the three circles is $\left(1^{2}+2^{2}+3^{2}\right) \pi=14 \pi \mathrm{cm}^{2}$.

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The vertices of a triangle whose sides measure 3, 4, and $5 \mathrm{~cm}$, are the centers of three circles that are pairwise externally tangent. What is the sum of the areas of these three circles?