(12) A small ball with a radius of 1 can move freely in all directions inside a regular tetrahedron container with inner wall edge length of $4 \sqrt{6}$. The area of the container's inner wall that the ball can never touch is $\qquad$
(12) As shown in Figure 1, consider the situation when the small ball is squeezed into a corner. Let the radius of the ball be $r$. Construct a plane $A_{1} B_{1} C_{1} \parallel$ plane $A B C$, tangent to the ball at point $D$. Then the center of the 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 $\triangle A_{1} B_{1} C_{1}$. Since
$$
\begin{aligned}
V_{P-A_{1} B_{1} C_{1}} & =\frac{1}{3} S_{\triangle A_{1} B_{1} C_{1}} \cdot P D \\
& =4 \cdot V_{O-A_{1} B_{1} C_{1}} \\
& =4 \cdot \frac{1}{3} \cdot S_{\triangle A_{1} B_{1} C_{1}} \cdot O D,
\end{aligned}
$$

it follows that $P D=4 O D=4 r$, thus
$$
P O=P D-O D=4 r-r=3 r .
$$

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

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

The area of the part of the face $P A B$ that the ball cannot touch (as shown in the shaded area in Figure 2) is
$$
\begin{aligned}
S_{\triangle P A B}-S_{\triangle P_{1} E F} & =\frac{\sqrt{3}}{4}\left(a^{2}-(a-2 \sqrt{6} r)^{2}\right) \\
& =3 \sqrt{2} a r-6 \sqrt{3} r^{2} .
\end{aligned}
$$

Given $r=1, a=4 \sqrt{6}$, we have
$$
S_{\triangle P A B}-S_{\triangle P_{1} E F}=24 \sqrt{3}-6 \sqrt{3}=18 \sqrt{3} .
$$

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