6. Given that the three vertices of $\triangle A B C$ are all on the parabola $y^{2}=2 p x(p>0)$, and the centroid of $\triangle A B C$ is exactly the focus of the parabola. If the equation of the line on which side $B C$ lies is $4 x+y$ $-20=0$, then $p=$ $\qquad$ .
6. $p=8$.

Let $A\left(\frac{y_{1}^{2}}{2 p}, y_{1}\right), B\left(\frac{y_{2}^{2}}{2 p}, y_{2}\right), C\left(\frac{y_{3}^{2}}{2 p}, y_{3}\right)$.
$$
\begin{array}{l}
\text { From }\left\{\begin{array}{l}
y^{2}=2 p x, \\
4 x+y-20=0
\end{array}\right. \\
\Rightarrow 2 y^{2}+p y-20 p=0 .
\end{array}
$$

From the given information,
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{y_{1}^{2}}{2 p}+\frac{y_{2}^{2}}{2 p}+\frac{y_{3}^{2}}{2 p}=\frac{3 p}{2}, \\
y_{1}+y_{2}+y_{3}=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
y_{2}+y_{3}=-\frac{p}{2}, \\
y_{2} y_{3}=-10 p, \\
y_{1}+y_{2}+y_{3}=0, \\
y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=3 p^{2} .
\end{array}\right.
\end{array}
$$

From equations (1) and (3), we get $y_{1}=\frac{p}{2}$.
Substituting into equation (4) and solving with (1) and (2) yields $p=8$.