6. Given three points $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$ on the unit circle $x^{2}+y^{2}=1$ that satisfy
$$
x_{1}+x_{2}+x_{3}=y_{1}+y_{2}+y_{3}=0 \text {. }
$$

then $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=$
6. $\frac{3}{2}$.

Let $x_{1}=\cos \alpha, x_{2}=\cos \beta, x_{3}=\cos \gamma$,
$$
y_{1}=\sin \alpha, y_{2}=\sin \beta, y_{3}=\sin \gamma .
$$

From the given conditions, the circumcenter, centroid, and orthocenter of $\triangle A B C$ coincide, indicating that it is an equilateral triangle.
$$
\begin{array}{l}
\text { Hence } \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma \\
=\frac{3}{2}+\frac{1}{2}(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma)=\frac{3}{2}, \\
\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma \\
=\frac{3}{2}-\frac{1}{2}(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma)=\frac{3}{2} .
\end{array}
$$