6. In the Cartesian coordinate system $x O y$, the ellipse $\Gamma: \frac{x^{2}}{4}+y^{2}=1$, point $L$ is in the third quadrant and $O L=3 \cdot \sqrt{2}$, and the angle between the negative half-axis of the $x$-axis and $O L$ is $\frac{\pi}{4}$. Two tangent lines $l_{1} 、 l_{2}$ are drawn through point $L$ to $\Gamma$. Let $l_{1} 、 l_{2}$ intersect the $y$-axis at points $P$ 、 $Q$, respectively. Then the value of $|O P| \cdot|O Q|$ is $\qquad$
Given by Zhou Xingjian, Question 9.
From the problem, we know $L(-3,-3)$, and the line $l: y+3=k(x+3)$ intersects with the ellipse. The combined equation is:
$$
\left(\frac{1}{4}+k^{2}\right) x^{2}+6 k(k-1) x+9(k-1)^{2}-1=0,
$$

If $l$ is tangent to the ellipse, then the discriminant of the above equation $\Delta=-5 k^{2}+18 k-8=0$. Let $l_{1}: y+3=k_{1}(x+3)$ and $l_{2}: y+3=k_{2}(x+3)$, then $k_{1}, k_{2}$ are the two solutions of $-5 k^{2}+18 k-8=0$. By Vieta's formulas, we have $k_{1}+k_{2}=\frac{18}{5}$ and $k_{1} k_{2}=\frac{8}{5}$. Therefore, $\left.|O P| \cdot|O Q|=\mid\left(3 k_{1}-3\right)\right)\left(3 k_{2}-3\right)|=9| k_{1} k_{2}-\left(k_{1}+k_{2}\right)+1 \mid=9$.