12. A small ball with a radius of 1 can move freely in all directions inside a regular tetrahedron container with an internal edge length of $4 \sqrt{6}$. Then the area of the container's inner wall that the ball can never touch is $\qquad$
$12.72 \sqrt{3}$.
As shown in Figure 3, consider the situation when the small ball is squeezed into a corner.
Let the radius of the small ball be $r$, and construct the plane $A_{1} B_{1} C_{1} /$ plane $A B C$, tangent to the small ball at point $D$. Then the center of the small ball $O$ is the center of the regular tetrahedron $P-A_{1} B_{1} C_{1}$, and $P O \perp$ plane $A_{1} B_{1} C_{1}$, with the foot of the perpendicular $D$ being the center of $\triangle A_{1} B_{1} C_{1}$.
By $V_{P-A_{1} B_{1} C_{1}}=\frac{1}{3} S_{\triangle A_{1} B_{1} C_{1}} \cdot P D=4 V_{O-A_{1} B_{1} C_{1}}$
$$
=4 \times \frac{1}{3} S_{\triangle A_{1} B_{1} C_{1}} \cdot O D \text {, }
$$

Thus, $P D=4 O D=4 r$.
Therefore, $P O=P D-O D=4 r-r=3 r$.
Let the point where the small ball touches the plane $P A B$ be $P_{1}$, and connect $O P_{1}$. Then
$$
P P_{1}=\sqrt{P O^{2}-O P_{1}^{2}}=\sqrt{(3 r)^{2}-r^{2}}=2 \sqrt{2} r .
$$

Consider the situation when the small ball is tangent to one of the faces of the regular tetrahedron (let's take $\triangle P A B$ for example). It is easy to see that the trajectory of the point where the small ball is closest to the edge on the plane $P A B$ is still a regular triangle, denoted as $\triangle P_{1} E F$ (as shown in Figure 4). Let the edge length of the regular tetrahedron be $a$, and draw $P_{1} M \perp P A$ at point $M$.
Since $\angle M P P_{1}=\frac{\pi}{6}$, we have
$$
P M=P P_{1} \cos \angle M P P_{1}=2 \sqrt{2} r \cdot \frac{\sqrt{3}}{2}=\sqrt{6} r .
$$

Thus, the side length of the small triangle
$$
P_{1} E=P A-2 P M=a-2 \sqrt{6} r \text {. }
$$

The area of the part that the small ball cannot touch on the plane $P A B$ (the shaded part in Figure 4) is
$$
\begin{array}{l}
S_{\triangle P A B}-S_{\triangle P_{1} E F}=\frac{\sqrt{3}}{4}\left[a^{2}-(a-2 \sqrt{6} r)^{2}\right] \\
=3 \sqrt{2} a r-6 \sqrt{3} r^{2} . \\
\text { Also, } r=1, a=4 \sqrt{6} \text {, then } \\
S_{\triangle A A B}-S_{\triangle P_{1} E F}=24 \sqrt{3}-6 \sqrt{3}=18 \sqrt{3} .
\end{array}
$$

By symmetry, and since the regular tetrahedron has 4 faces, the total area of the container's inner wall that the small ball cannot touch is $72 \sqrt{3}$.