3. Two spheres with a radius of 1 are externally tangent to each other, and both are internally tangent to a cylindrical surface with a radius of 1. Another smaller sphere is externally tangent to both of these spheres and internally tangent to the cylindrical surface. A plane passing through the center of the smaller sphere and the center of one of the larger spheres intersects the cylindrical surface to form an ellipse. The maximum possible value of the eccentricity \( e \) of this ellipse is \(\qquad\)
3. $\frac{4}{5}$.

As shown in the figure, let the two spheres $O_{1}$ and $O_{2}$ with radius 1 be externally tangent at point $S$, and the smaller sphere $O_{3}$ be internally tangent to the cylindrical surface at point $T$. The radius of the smaller sphere $O_{3}$ is $r$. By the Pythagorean theorem, we have
$$
1=T S=T O_{3}+O_{3} S=r+\sqrt{(1+r)^{2}-1^{2}} .
$$

Rearranging and squaring both sides, we solve to get $r=\frac{1}{4}$, which means the radius of the smaller sphere is $\frac{1}{4}$.

Let the line $O_{1} O_{3}$ intersect the cylindrical surface at points $A$ and $B$. Then $A B$ is the major axis of the ellipse, and its length is
$$
\begin{aligned}
2 a & =A B \\
& =2 \cdot O_{1} A \\
& =2 \cdot \frac{1}{\sin \angle O_{3} O_{1} S} \\
& =2 \cdot \frac{1}{\frac{1-\frac{1}{4}}{1+\frac{1}{4}}} \\
& =\frac{10}{3},
\end{aligned}
$$
$\boldsymbol{T}$
Thus, $a=\frac{5}{3}$.
A
Let the length of the minor axis of the ellipse be $b$, and the endpoints of the minor axis be $C$ and $D$. Clearly, $C D$ passes through point $O_{1}$, and the minimum length of $C D$ is 2, so the minimum value of $b$ is 1 (when $C D$ is perpendicular to the plane $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$).
$$
\begin{aligned}
e & =\frac{c}{a}=\frac{\sqrt{a^{2}-b^{2}}}{a} \\
& \geqslant \frac{\sqrt{\left(\frac{5}{3}\right)^{2}-1^{2}}}{\frac{5}{3}}=\frac{4}{5}
\end{aligned}
$$

Therefore, the maximum value of the eccentricity $e$ is $\frac{4}{5}$.