Six. (16 points) As shown in Figure 3, there is a fixed point $P$ inside $\angle M A N$. It is known that $\tan \angle M A N=3, P$ is at a distance $P D=12$ from the line $A N$, and $A D=30$. A line through $P$ intersects $A N$ and $A M$ at points $B$ and $C$ respectively. Find the minimum area of $\triangle A B C$.

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Six, Solution 1
As shown in Figure 7, let $D B=$
$x$, draw $C E \perp A B$,
$P F / / C A$.
Let $A E=t$. Then
$$
\begin{array}{l}
\tan \angle M A N=3 \\
\Rightarrow C E=3 t .
\end{array}
$$

Similarly, by $P D=12 \Rightarrow F D=4$.
By $\triangle A B C \backsim \triangle F B P$, we get $\frac{A B}{C E}=\frac{F B}{P D}$, i.e.,
$$
\frac{30+x}{3 t}=\frac{4+x}{12} \text {. }
$$

Thus, $t=\frac{4(30+x)}{4+x}$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} A B \cdot C E$
$=\frac{1}{2}(30+x) \times \frac{12(30+x)}{4+x}$
$=6 \times \frac{(30+x)^{2}}{4+x}$.
Let $m=\frac{(30+x)^{2}}{4+x}$.
Transforming into a quadratic equation in $x$ gives
$$
x^{2}+(60-m) x+900-4 m=0 \text {. }
$$

Since $x$ is a real number, we have
$$
\Delta=(60-m)^{2}-4(900-4 m) \geqslant 0 \text {, }
$$

i.e., $m(m-104) \geqslant 0$.
Thus, $m \leqslant 0$ or $m \geqslant 104$.
Therefore, when $m=104$, we have
$\left(S_{\triangle B B C}\right)_{\text {min }}=6 \mathrm{~m}=624$.
By $\frac{(30+x)^{2}}{4+x}=104$, solving for $x$ gives $x=22$.
Solution 2 can prove that when the line through point $P$ satisfies $P B=P C$, the area of $\triangle A B C$ is minimized.

In fact, let $B_{1} C_{1}$ be any line through point $P$, forming $\triangle A B_{1} C_{1}$.
As shown in Figure 8, draw $C F$
$/ / B_{1} B$. Then
$\triangle P C F \cong \triangle P B B_{1}$.
Thus, $S_{\triangle B B C}$
$$
\begin{array}{l}
=S_{\triangle P C F}+ \\
S_{\text {quadrilateral } 1 B_{1}, P C} \\
<S_{\triangle B_{1} C_{1}} \text {. } \\
\end{array}
$$

According to the above geometric conclusion, the calculation is as follows.
Since $P D=12$, we have $C E=24$.
Since $\tan \angle M A N=3$, we have $A E=8$.
By $E D=B D$, and $E D=A D-A E=22$, we get $B D=22$.
Thus, $A B=A D+B D=52$.
$\left(S_{\triangle A B C}\right)_{\text {min }}=\frac{1}{2} A B \cdot C E=624$.