11. (1999 Vietnam Mathematical Olympiad) Let $a, b, c$ be positive real numbers, and $abc + a + c = b$. Determine $p=$

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11. From the given condition, we have $a+c=(1-a c) b$. Clearly, $1-a c \neq 0$, so $b=\frac{a+c}{1-a c}$. Let $a=\tan \alpha, b=\tan \beta, c=\tan \gamma$, where $\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$. Then, $\tan \beta=\frac{\tan \alpha+\tan \gamma}{1-\tan \alpha \cdot \tan \gamma}=\tan (\alpha+\gamma)$.

Since $\beta, \alpha+\gamma \in(0, \pi)$, it follows that $\beta=\alpha+\gamma$. Therefore, 
\[ p=\frac{2}{1+\tan ^{2} \alpha}-\frac{2}{1+\tan ^{2} \beta}+\frac{3}{1+\tan ^{2} \gamma}=2 \cos ^{2} \alpha-2 \cos ^{2}(\alpha+\gamma)+3 \cos ^{2} \gamma \]
\[ = (1+\cos 2 \alpha)-[1+\cos (2 \alpha+2 \gamma)]+3 \cos ^{2} \gamma \]
\[ = 2 \sin \gamma \cdot \sin (2 \alpha+\gamma)+3 \cos ^{2} \gamma \leqslant 2 \sin \gamma+3\left(1-\sin ^{2} \gamma\right) \]
\[ = -3\left(\sin \gamma-\frac{1}{3}\right)^{2}+\frac{10}{3} \leqslant \frac{10}{3} \]
The equality holds when $2 \alpha+\gamma=\frac{\pi}{2}$ and $\sin \gamma=\frac{1}{3}$. Hence, when $a=\frac{\sqrt{2}}{2}, b=\sqrt{2}, c=\frac{\sqrt{2}}{4}$, $p_{\text {max }}=\frac{10}{3}$.