7. The maximum value of the projection area of a rectangular prism with edge lengths $6,6,8$ on plane $\alpha$ is $\qquad$
7. $12 \sqrt{41}$.

Lemma If three planes are pairwise perpendicular, and a fourth plane in space forms angles $x, y, z$ with these three planes, then $\cos ^{2} x+\cos ^{2} y+\cos ^{2} z=1$.
Proof Let the projection of $O$ on plane $ABC$ be $D$. Then the normal vector of plane $OAB$ is $\overrightarrow{OC}$, and the normal vector of plane $ABC$ is $\overrightarrow{DO}$. Hence, $x$ is the angle formed by $OD$ and $OC$.
Thus, $\cos x=\frac{OD}{OC}$.
Similarly, $\cos y=\frac{OD}{OB}, \cos z=\frac{OD}{OA}$.
By a familiar conclusion
$$
\begin{array}{l}
\frac{1}{OD^{2}}=\frac{1}{OA^{2}}+\frac{1}{OB^{2}}+\frac{1}{OC^{2}} \\
\Rightarrow \cos ^{2} x+\cos ^{2} y+\cos ^{2} z \\
\quad=OD^{2}\left(\frac{1}{OA^{2}}+\frac{1}{OB^{2}}+\frac{1}{OC^{2}}\right)=1 .
\end{array}
$$

The lemma is proved.
The projection area of a rectangular prism on plane $\alpha$ is the sum of the projection areas of the three adjacent faces. Let the half areas of the three adjacent faces be $s_{1}, s_{2}, s_{3}$. Then by the spatial Pythagorean theorem, we have $S^{2}=s_{1}^{2}+s_{2}^{2}+s_{3}^{2}$, where $S$ represents the area of the triangle formed by the diagonals of these three faces, and let the cosines of the angles formed by these three planes with plane $\alpha$ be $\cos x, \cos y, \cos z$, hence by the projection theorem and the spatial Pythagorean theorem, we have
$$
\cos ^{2} x+\cos ^{2} y+\cos ^{2} z=1 \text {. }
$$

Thus, the sum of the projection areas
$$
x=2\left(s_{1} \cos x+s_{2} \cos y+s_{3} \cos z\right) \text {. }
$$

Therefore, by the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
x \leqslant 2 \sqrt{\left(s_{1}^{2}+s_{2}^{2}+s_{3}^{2}\right)\left(\cos ^{2} x+\cos ^{2} y+\cos ^{2} z\right)} \\
=2 S=12 \sqrt{41} .
\end{array}
$$