6. 27 .

Let the three vertices of $\triangle A B C$ have coordinates
$$
A\left(x_{A}, y_{A}\right), B\left(x_{B}, y_{B}\right), C\left(x_{C}, y_{C}\right) \text {. }
$$

The focus of the parabola $y^{2}=4 \sqrt{2} x$ is $F(\sqrt{2}, 0)$.
From the given conditions, we have
$$
\left\{\begin{array}{l}
x_{A}+x_{B}+x_{C}=3 \sqrt{2}, \\
y_{A}+y_{B}+y_{C}=0 .
\end{array}\right.
$$

Then $y_{A}^{2}+y_{B}^{2}+y_{C}^{2}$
$$
\begin{array}{l}
=4 \sqrt{2}\left(x_{A}+x_{B}+x_{C}\right)=24, \\
y_{B} y_{C}+y_{C} y_{A}+y_{A} y_{B} \\
=\frac{1}{2}\left(y_{A}+y_{B}+y_{C}\right)^{2}-\frac{1}{2}\left(y_{A}^{2}+y_{B}^{2}+y_{C}^{2}\right) \\
=-12 .
\end{array}
$$

Therefore, $y_{A}^{4}+y_{B}^{4}+y_{C}^{4}$
$$
\begin{aligned}
= & \left(y_{A}^{2}+y_{B}^{2}+y_{C}^{2}\right)^{2}-2\left(y_{B} y_{C}+y_{C} y_{A}+y_{A} y_{B}\right)^{2}+ \\
& 4 y_{A} y_{B} y_{C}\left(y_{A}+y_{B}+y_{C}\right) \\
= & 24^{2}-2 \times 12^{2}=288 .
\end{aligned}
$$

Accordingly,
$$
\begin{array}{l}
|F A|^{2}+|F B|^{2}+|F C|^{2} \\
=\left(x_{A}-\sqrt{2}\right)^{2}+y_{A}^{2}+\left(x_{B}-\sqrt{2}\right)^{2}+ \\
y_{B}^{2}+\left(x_{C}-\sqrt{2}\right)^{2}+y_{C}^{2} \\
= \frac{y_{A}^{4}+y_{B}^{4}+y_{C}^{4}}{(4 \sqrt{2})^{2}}-2 \sqrt{2}\left(x_{A}+x_{B}+x_{C}\right)+ \\
3 \times 2+y_{A}^{2}+y_{B}^{2}+y_{C}^{2} \\
= 9-12+6+24=27 .
\end{array}
$$
6. If the centroid of the inscribed $\triangle A B C$ of the curve $y^{2}=4 \sqrt{2} x$ is its focus $F$. Then
$$
|F A|^{2}+|F B|^{2}+|F C|^{2}=
$$
$\qquad$