8. If $a, b, c$ are distinct integers, then
$$
3 a^{2}+2 b^{2}+4 c^{2}-a b-3 b c-5 c a
$$
the minimum value is . $\qquad$
8. 6 .

Notice that,
$$
\begin{array}{l}
3 a^{2}+2 b^{2}+4 c^{2}-a b-3 b c-5 c a \\
=\frac{1}{2}(a-b)^{2}+\frac{3}{2}(b-c)^{2}+\frac{5}{2}(c-a)^{2} .
\end{array}
$$

Since \(a, b, c\) are distinct integers, when \(a-b=2, a-c=1, c-b=1\), the original expression achieves its minimum value of 6.