6. Let $|\overrightarrow{A B}|=10$. If a point $P$ on the plane satisfies that for any $t \in \mathbf{R}$, $|\overrightarrow{A P}-t \overrightarrow{A B}| \geqslant 3$, then the minimum value of $\overrightarrow{P A} \cdot \overrightarrow{P B}$ is $\qquad$, and at this time, $|\overrightarrow{P A}+\overrightarrow{P B}|=$ $\qquad$.
6. $-16,6$.

From $|\overrightarrow{A P}-t \overrightarrow{A B}| \geqslant 3$, we know that the distance from point $P$ to line $A B$ is 3.
Let the midpoint of $A B$ be $O$. Then
$$
\begin{aligned}
& \overrightarrow{P A} \cdot \overrightarrow{P B} \\
& =\frac{1}{4}\left((\overrightarrow{P A}+\overrightarrow{P B})^{2}-(\overrightarrow{P A}-\overrightarrow{P B})^{2}\right) \\
= & \frac{1}{4}\left((2 \overrightarrow{P O})^{2}-10^{2}\right) \\
\geqslant & \frac{1}{4}(36-100) \\
= & -16 .
\end{aligned}
$$

At this point, $|\overrightarrow{P A}+\overrightarrow{P B}|=6$.