One, Solution: As shown in the figure, construct $\square A B M C$ and $\square A D N C$, then
$$
\begin{array}{l}
\overrightarrow{D M}^{2}+\overrightarrow{D A}^{2}-\left(\overrightarrow{D B}^{2}+\overrightarrow{D C}^{2}\right) \\
=(\overrightarrow{D A}+\overrightarrow{A M})^{2}+\overrightarrow{D A}^{2}-(\overrightarrow{D A}+\overrightarrow{A B})^{2}-(\overrightarrow{D A}+\overrightarrow{A C})^{2} \\
=2(\overrightarrow{D A} \cdot \overrightarrow{A M}-\overrightarrow{D A} \cdot \overrightarrow{A B}-\overrightarrow{D A} \cdot \overrightarrow{A C})+\overrightarrow{A M}^{2}-\overrightarrow{A B}^{2}-
\end{array}
$$
$\overrightarrow{A C}^{2}$,
Since $A B M C$ is a parallelogram and $\angle B A C=60^{\circ}$,
we have $\overrightarrow{A M}=\overrightarrow{A B}+\overrightarrow{A C}$, thus $\overrightarrow{D A} \cdot \overrightarrow{A M}-\overrightarrow{D A} \cdot \overrightarrow{A B}-\overrightarrow{D A} \cdot \overrightarrow{A C}=0$,
and $\overrightarrow{A M}^{2}-\overrightarrow{A B}^{2}-\overrightarrow{A C}^{2}=2 \overrightarrow{A B} \cdot \overrightarrow{A C}=A B \cdot A C$.
Therefore, $\overrightarrow{D M}^{2}+64-(64+36)=A B \cdot A C$, and it is easy to see that $A B \cdot A C=C M \cdot D N$.
By Ptolemy's theorem, $C M \cdot D N \leqslant C D \cdot M N+C N \cdot D M=48+8 D M$,
so $D M^{2}-8 D M-84 \leqslant 0$, solving this gives $D M \leqslant 14$, with equality if and only if $C, D, M, N$ are concyclic.
The necessary and sufficient condition for $C, D, M, N$ to be concyclic is $\angle C D N=\angle C M N$, i.e., $\angle A C D=\angle A B D$,
so when $\angle A C D=\angle A B D$, $S_{\triangle A B C}=\frac{\sqrt{3}}{4} A B \cdot A C=\frac{\sqrt{3}}{4}\left(D M^{2}-36\right)=40 \sqrt{3}$, which is the required maximum value.
One, (50 points) In $\triangle A B C$, $\angle A=60^{\circ}$, and $D$ is a point on the plane of $\triangle A B C$ such that $D A=8, D B=$ $8, D C=6$. Find the maximum area of $\triangle A B C$.