14. A. As shown in Figure 7, in $\triangle ABC$, it is given that $\angle BAC=60^{\circ}, AB=$ $2AC$, point $P$ is inside $\triangle ABC$, and $PA=\sqrt{3}, PB=$ $5, PC=2$. Find the area of $\triangle ABC$.
14. A. As shown in Figure 11, construct
$\triangle A B Q$ such that
$$
\begin{array}{l}
\angle Q A B=\angle P A C, \\
\angle A B P=\angle A C P .
\end{array}
$$

Then $\triangle A B Q$
$$
\backsim \triangle A C P \text {. }
$$

Since $A B=2 A C$, the similarity ratio is 2.
Thus, $A Q=2 A P=2 \sqrt{3}, B Q=2 C P=4$.
Therefore, $\angle Q A P=\angle Q A B+\angle B A P$
$$
=\angle P A C+\angle B A P=\angle B A C=60^{\circ} \text {. }
$$

Given $A Q: A P=2: 1$, we know $\angle A P Q=90^{\circ}$.
Thus, $P Q=\sqrt{3} A P=3$.
So, $B P^{2}=25=B Q^{2}+P Q^{2}$.
Hence, $\angle B Q P=90^{\circ}$.
Then $A B^{2}=P Q^{2}+(A P+B Q)^{2}=28+8 \sqrt{3}$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} A B \cdot A C \sin 60^{\circ}$
$$
=\frac{\sqrt{3}}{8} A B^{2}=\frac{6+7 \sqrt{3}}{2} \text {. }
$$