Question 95, Given 0 as the circumcenter of $\triangle A B C$, satisfying $\overrightarrow{C O}=t \cdot \overrightarrow{C A}+\left(\frac{1}{2}-\frac{3 t}{4}\right) \cdot \overrightarrow{C B}(t \in R \backslash\{0\})$, if $|A B|=3$, try to find the maximum value of the area of $\triangle A B C$.
Problem 95, Solution Method One: As shown in the figure, draw $O D \perp B C$ at $D$, then $D$ is the midpoint of $B C$. Draw $A E \perp B C$ at $E$, and take the midpoint of $A B$ as $F$.
According to the conditions:
$$
\begin{array}{l}
\overrightarrow{\mathrm{CO}} \cdot \overrightarrow{\mathrm{CB}}=\mathrm{t} \cdot \overrightarrow{\mathrm{CA}} \cdot \overrightarrow{\mathrm{CB}}+\left(\frac{1}{2}-\frac{3 \mathrm{t}}{4}\right) \cdot \overrightarrow{\mathrm{CB}} \cdot \overrightarrow{\mathrm{CB}} \\
\Rightarrow|\mathrm{CD}| \cdot|\mathrm{CB}|=\mathrm{t} \cdot|\mathrm{CE}| \cdot|\mathrm{CB}|+\left(\frac{1}{2}-\frac{3 t}{4}\right) \cdot|\mathrm{CB}|^{2} \\
\Rightarrow|\mathrm{CE}|=\frac{3|\mathrm{CB}|}{4}
\end{array}
$$

Thus, we know $S_{\triangle A B C}=4 S_{\triangle A B E} \leq 4 \cdot \frac{A B \cdot E F}{2}=4 \cdot \frac{3 \times \frac{3}{2}}{2}=9$.
Solution Method Two: As shown in the figure, draw $O D \perp B C$ at $D$, then $D$ is the midpoint of $B C$. Draw $A E \perp B C$ at $E$, and take the midpoint of $A B$ as $F$. According to the conditions:
$$
\begin{array}{l}
-\overline{\mathrm{OC}}=\mathrm{t} \cdot(\overrightarrow{\mathrm{OA}}-\overrightarrow{\mathrm{OC}})+\left(\frac{1}{2}-\frac{3 \mathrm{t}}{4}\right) \cdot(\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OC}}) \\
\Rightarrow \mathrm{t} \cdot \overrightarrow{\mathrm{OA}}+\left(\frac{1}{2}-\frac{3 \mathrm{t}}{4}\right) \cdot \overrightarrow{\mathrm{OB}}+\left(\frac{1}{2}-\frac{\mathrm{t}}{4}\right) \cdot \overrightarrow{\mathrm{OC}}=\overrightarrow{0}
\end{array}
$$

According to the Mercedes-Benz theorem: $\mathrm{S}_{\triangle B O C}: S_{\triangle A O C}: S_{\triangle A O B}=\mathrm{t}:\left(\frac{1}{2}-\frac{3 \mathrm{t}}{4}\right):\left(\frac{1}{2}-\frac{\mathrm{t}}{4}\right)$, hence
$S_{\triangle A O B}-S_{\triangle A O C}=\frac{S_{\triangle B O C}}{2}$
$\Rightarrow S_{\text {quadrilateral } A O D B}-S_{\triangle A O D C}=\frac{S_{\triangle B O C}}{2}$
$\Rightarrow S_{\triangle A O D}=\frac{S_{\triangle B O C}}{4}$
$\Rightarrow S_{\triangle E O D}=\frac{S_{\triangle B O C}}{4}$
$\Rightarrow|D E|=\frac{|B C|}{2}$
$\Rightarrow|B E|=\frac{|B C|}{4}$
Thus, we know $S_{\triangle A B C}=4 S_{\triangle A B E} \leq 4 \cdot \frac{A B \cdot E F}{2}=4 \cdot \frac{3 \times \frac{3}{2}}{2}=9$.
Solution Method Three: As shown in the figure, draw $O D \perp B C$ at $D$, then $D$ is the midpoint of $B C$. Take the midpoint of $B D$ as $E$, and the midpoint of $A B$ as $F$. According to the conditions:
$$
\begin{array}{l}
\overrightarrow{\mathrm{CO}}=\mathrm{t} \cdot \overrightarrow{\mathrm{CA}}+\left(\frac{1}{2}-\frac{3 \mathrm{t}}{4}\right) \cdot \overrightarrow{\mathrm{CB}}=\frac{\overline{\mathrm{CB}}}{2}+\mathrm{t}\left(\overrightarrow{\mathrm{CA}}-\frac{3 \overline{\mathrm{CB}}}{4}\right)=\overrightarrow{\mathrm{CD}}+\mathrm{t}(\overrightarrow{\mathrm{CA}}-\overrightarrow{\mathrm{CE}}) \\
\Rightarrow \overrightarrow{\mathrm{DO}}=\mathrm{t} \cdot \overrightarrow{\mathrm{EA}} \\
\Rightarrow \mathrm{AE} \perp \mathrm{BC}
\end{array}
$$

Thus, we know $S_{\triangle A B C}=4 S_{\triangle A B E} \leq 4 \cdot \frac{A B \cdot E F}{2}=4 \cdot \frac{3 \times \frac{3}{2}}{2}=9$.