$6 . A 、 B 、 C 、 D$ Four football teams are playing a round-robin tournament (i.e., any two teams will play against each other). After several matches, the match situations of teams $A 、 B 、 C$ are as follows:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline & Matches & Wins & Losses & Draws & Goals Scored & Goals Conceded \\
\hline$A$ Team & 2 & 0 & 2 & 0 & 3 & 6 \\
\hline$B$ Team & 2 & 1 & 0 & 1 & 4 & 3 \\
\hline$C$ Team & 3 & 2 & 0 & 1 & 2 & 0 \\
\hline \begin{tabular}{c}
$D$ Team \\
(Answer)
\end{tabular} & & & & & & \\
\hline
\end{tabular}

Question: Please fill in the correct numbers in the above blanks.
6. Since Team C has played 3 matches, we start with Team C. Team C won 2 matches and drew 1, so the match results of Team C against other teams can only be 1:0 (win), 1:0 (win), and 0:0 (draw). Since Team B drew 1 match, the match result between Team C and Team B is 0:0, and the match results between Team C and Team A, and Team C and Team D are both 1:0. (See Figure 2)
Next, consider Teams A and B. Since both Teams A and B have only played 2 matches, there are only two possibilities: (1) A vs. B; (2) A vs. D and B vs. D. For the first scenario, based on the goals conceded by Team A, Team B should have scored 5 goals, but Team B only scored 4 goals, which is a contradiction. Therefore, only the second scenario is possible: A vs. D and B vs. D. The 6 goals conceded by Team A, apart from the 1 goal scored by Team C, must have all been scored by Team D. Therefore, the match result between Team A and Team D is 3:5, and the match result between Team B and Team D is 4:3 (See Figure 3).

Thus, Team D has played 3 matches, won 1, lost 2, scored 8 goals, and conceded 8 goals.