6. Given $F_{1} 、 F_{2}$ are the foci of the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0)$, $P$ is a point on $\Gamma$, the perimeter of $\triangle P F_{1} F_{2}$ is 6, and the minimum value of $\frac{4}{a}+\frac{1}{c}$ is 3. A line passing through the point $Q(-4,0)$ intersects $\Gamma$ at two distinct points $A 、 B$. Then the range of $|Q A||Q B|$ is
6. $\left(\frac{45}{4}, 12\right]$.

The perimeter of $\triangle P F_{1} F_{2}$ is
$$
\begin{array}{l}
2 a+2 c=6 \Rightarrow a+c=3 . \\
\text { By }\left(\frac{4}{a}+\frac{1}{c}\right)(a+c) \geqslant(2+1)^{2} \\
\Rightarrow \frac{4}{a}+\frac{1}{c} \geqslant 3,
\end{array}
$$

When $a=2, c=1$, the equality holds.
Thus, the ellipse $\Gamma: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$.
Let the equation of line $Q A$ be $x=m y-4$. Solving together, we get
$$
\left(\frac{1}{3}+\frac{m^{2}}{4}\right) y^{2}-2 m y+3=0 \text {. }
$$

By $\Delta=(2 m)^{2}-4 \times 3\left(\frac{1}{3}+\frac{m^{2}}{4}\right) \geqslant 0$
$\Rightarrow m \geqslant 2$ or $m<-2$.
Then $|Q A||Q B|=\left|y_{1} y_{2}\right|\left(\sqrt{1+m^{2}}\right)^{2}$
$$
=\frac{3\left(1+m^{2}\right)}{\frac{m^{2}}{4}+\frac{1}{3}} \text {. }
$$

Substituting $m \geqslant 2$ or $m<-2$ into the above expression, we get the range of $|Q A||Q B|$ is $\left(\frac{45}{4}, 12\right]$.