使用左连接：

SELECT
     I.id
    ,T.title
FROM images AS i
LEFT JOIN articles_images AS ai ON I.id = ai.image_id
LEFT JOIN topics AS t on ai.article_id = t.id
WHERE i.vender_id = 'model' 
ORDER BY
    i.id DESC
LIMIT 10

结果与您要求的结果略有不同：

+----+-------+
| id | title |
+----+-------+
|  8 | NULL  |
|  7 | NULL  |
|  5 | AAA   |
|  5 | CCC   |
|  4 | DDD   |
|  3 | BBB   |
|  2 | BBB   |
|  1 | NULL  |
+----+-------+

对于重新修改的问题（包括空值）：

SELECT
     I.id
    ,max(T.title) as title
FROM images AS i
LEFT JOIN articles_images AS ai ON I.id = ai.image_id
LEFT JOIN topics AS t on ai.article_id = t.id
WHERE i.vender_id = 'model' 
GROUP BY
     I.id
;

+----+-------+
| id | title |
+----+-------+
|  1 | NULL  |
|  2 | BBB   |
|  3 | BBB   |
|  4 | DDD   |
|  5 | CCC   |
|  7 | NULL  |
|  8 | NULL  |
+----+-------+

请参见：https://rextester.com/lcobw94449
为了保证您使用的是最高的文章\u，如果我建议：

SELECT
     I.id
    ,T.title
FROM ximages AS i
LEFT JOIN (
            SELECT image_id, max(article_id) article_id
            FROM xarticles_images
            GROUP BY image_id
          ) AS ai ON I.id = ai.image_id
LEFT JOIN xtopics AS t on ai.article_id = t.id
WHERE i.vender_id = 'model' 
ORDER BY
    i.id DESC
;

使用左连接

SELECT
        i.`id`,
        t.`title`
    FROM
        `images` AS i left join `topics` AS  t on  i.`vender_id` = 'model' 
         and t.`id`=i.article_id