6. As shown in Figure 2, let $P$ be a point inside the equilateral $\triangle ABC$ with side length 12. Draw perpendiculars from $P$ to the sides $BC$, $CA$, and $AB$, with the feet of the perpendiculars being $D$, $E$, and $F$ respectively. Given that $PD: PE: PF = 1: 2: 3$. Then, the area of quadrilateral $BDPF$ is
$$
6.11 \sqrt{3} \text {. }
$$

For the equilateral $\triangle ABC$ with a height of $12 \times \frac{\sqrt{3}}{2}=6 \sqrt{3}$, then
$$
\begin{array}{l}
P D+P E+P F=6 \sqrt{3} . \\
\text { Also, } P D: P E: P F=1: 2: 3, \text { so, } \\
P D=\sqrt{3}, P E=2 \sqrt{3}, P F=3 \sqrt{3} .
\end{array}
$$

As shown in Figure 7, draw
$P S / / B C$, intersecting $A B$ at $S$.
Since $\angle F S P=\angle B$ $=60^{\circ}$, therefore,
$$
\begin{array}{l}
F S=P F \cot 60^{\circ}=3, \\
P S=6, \\
B D=P S+P D \cot 60^{\circ}=7 .
\end{array}
$$

Thus, $S_{\text {quadrilateral BDPF }}=S_{\triangle P S}+S_{\text {trapezoid BDPS }}$
$$
=\frac{1}{2} \times 3 \times 3 \sqrt{3}+\frac{6+7}{2} \times \sqrt{3}=11 \sqrt{3} .
$$