Question 93: Planar vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy: $|\vec{a}|=|\vec{b}| \neq 0, \vec{a} \perp \vec{b},|\vec{c}|=2 \sqrt{2},|\vec{c}-\vec{a}|=1$, try to find the maximum possible value of $|\vec{a}+\vec{b}-\vec{c}|$.
Question 93, Solution: As shown in the figure, let $\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{c}}, \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\overline{\mathrm{OD}}$, then quadrilateral $0 \mathrm{ADB}$ is a square, $|\vec{c}-\vec{a}|=|A C|=1$, and $|\vec{a}+\vec{b}-\vec{c}|=|C D|$. Let the side length of the square $O A D B$ be $x$, then $|O D|=$ $\sqrt{2} x$. According to Ptolemy's theorem, we have:
$$
\begin{array}{l}
|C D| \cdot|O A| \leq|O D| \cdot|A C|+|A D| \cdot|O C| \\
\Rightarrow x \cdot|C D| \leq \sqrt{2} x \cdot 1+x \cdot 2 \sqrt{2} \\
\Rightarrow|C D| \leq 3 \sqrt{2}
\end{array}
$$

Therefore, the maximum possible value of $|\vec{a}+\vec{b}-\vec{c}|$ is $3 \sqrt{2}$.