6. As shown in the figure, let the parabola $y=a x^{2}+b x+c(a, b, c \in \mathbb{R}, a>0)$ intersect the $x$-axis at two distinct points $A, B$, with the vertex at $C$. The tangents to the parabola at $A$ and $B$ intersect at point $P$. Let $S_{\triangle A P B}=S_{1}$, the area enclosed by the parabolic arc $\overparen{A C B}$ and the line segment $A B$ be $S_{2}$, and $S_{\triangle A C B}=S_{3}$. Then $S_{1}: S_{2}: S_{3}=$ $\qquad$
Solution: $6: 4: 3$.
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)\left(x_{1}<x_{2}\right)$
$$
\Delta=b^{2}-4 a c,\left|x_{1}-x_{2}\right|^{2}=\frac{\Delta}{a^{2}}
$$

From $y^{\prime}=2 a x+b$ we get $k_{P A}=2 a x_{1}+b$.
Thus, the equation of line $P A$ is $l_{P A}: y=\left(2 a x_{1}+b\right)\left(x-x_{1}\right)$
Since the x-coordinate of $P$ is $\frac{x_{1}+x_{2}}{2}$, the y-coordinate is
$$
\begin{array}{c}
\frac{\left(2 a x_{1}+b\right)\left(x_{2}-x_{1}\right)}{2}=\frac{-a\left(x_{2}-x_{1}\right)^{2}}{2}=-\frac{\Delta}{2 a} \\
P\left(-\frac{b}{2 a},-\frac{\Delta}{2 a}\right)
\end{array}
$$

Therefore, $S_{1}=\frac{1}{2}\left|x_{1}-x_{2}\right|\left|-\frac{\Delta}{2 a}\right|=\frac{\Delta \sqrt{\Delta}}{4 a^{2}}, S_{3}=\frac{1}{2}\left|x_{1}-x_{2}\right|\left|-\frac{\Delta}{4 a}\right|=\frac{\Delta \sqrt{\Delta}}{8 a^{2}}$
Also,
$$
\begin{aligned}
S_{2}=\mid \int_{x_{1}}^{x_{2}}\left(a x^{2}+\right. & b x+c) \left.\mathrm{d} x|=| \frac{1}{3} a\left(x_{2}^{3}-x_{1}^{3}\right)+\frac{1}{2} b\left(x_{2}^{2}-x_{1}^{2}\right)+c\left(x_{2}-x_{1}\right) \right\rvert\, \\
& =\left|x_{1}-x_{2}\right|\left|\frac{1}{3} a\left(\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}\right)+\frac{1}{2} b\left(x_{1}+x_{2}\right)+c\right|=\frac{\sqrt{\Delta}}{a}\left|\frac{2}{3} c-\frac{b^{2}}{6 a}\right| \\
& =\frac{\Delta \sqrt{\Delta}}{6 a^{2}}
\end{aligned}
$$

Hence, $S_{1}: S_{2}: S_{3}=6: 4: 3$