7. A small sphere with radius $r$ and a large sphere with radius $R$ are freely moving in all directions inside a regular tetrahedron container with internal edge length $l$. If
$$
r=1, R=2, l=\frac{14 \sqrt{6}}{3},
$$

then the area of the container's inner wall that the small sphere can never touch is . $\qquad$
7. $88 \sqrt{3}+\frac{128 \pi}{9}$.

Let the edge length of a regular tetrahedron be $l_{0}$ and the radius of its inscribed sphere be $r_{0}$. It is known that $l_{0}=2 \sqrt{6} r_{0}$, and the distance from the points where the inscribed sphere touches the faces to the edges is $\sqrt{2} r_{0}$.

First, consider the case where only the small sphere is inside the regular tetrahedron container, and calculate the area of the container's inner wall that the small sphere cannot touch.
The area of the part on one face that the small sphere cannot touch
(as shown in Figure 3, shaded area)
is
$$
\frac{\sqrt{3}}{4}\left[l^{2}-(l-2 \sqrt{6} r)^{2}\right]=22 \sqrt{3} .
$$

Next, consider the scenario where both the large and small spheres are inside the regular tetrahedron container.

If both spheres are tangent to one face of the regular tetrahedron, the distance between the points of tangency is no less than
$$
\begin{array}{l}
\sqrt{(R+r)^{2}-(R-r)^{2}}=2 \sqrt{R r} . \\
\text { When } 2 \sqrt{2} r+2 \sqrt{R r}+\sqrt{2} R \leqslant \frac{\sqrt{3}}{2} l \text {, the small }
\end{array}
$$

sphere can be squeezed into one corner of the regular tetrahedron, allowing the large sphere to move from one corner to another;

When $2 \sqrt{2} R+2 \sqrt{R r}+\sqrt{2} r \leqslant \frac{\sqrt{3}}{2} l$, the large sphere can be squeezed into one corner of the regular tetrahedron, allowing the small sphere to move from one corner to another.
$$
\begin{array}{l}
\text { Combining } R=2, r=1, l=\frac{14 \sqrt{6}}{3} \text {, we have } \\
\frac{\sqrt{3}}{2} l=2 \sqrt{2} R+2 \sqrt{R r}+\sqrt{2} r \\
>2 \sqrt{2} r+2 \sqrt{R r}+\sqrt{2} R .
\end{array}
$$

Therefore, both the large and small spheres can move from one corner of the regular tetrahedron to another.

When the large sphere is squeezed into the top corner, the area of the circular part in the center of the base that the small sphere cannot touch is $\pi r_{1}^{2}$, where
$$
r_{1}^{2}=(R+r)^{2}-\left(\frac{\sqrt{6}}{3} l-3 R-r\right)^{2}=\frac{32}{9} .
$$

In summary, the required area is
$$
4\left(22 \sqrt{3}+\frac{32 \pi}{9}\right)=88 \sqrt{3}+\frac{128 \pi}{9} .
$$