We want a remainder of $ 10 $ when divided by both $ 11 $ and $ 13 $. The least common multiple of $ 11 $ and $ 13 $ is $ 143 $. We add $ 10 $ to the number such that the remainder would be $ 10 $ when divided by $ 11 $ and $ 13 $ so we get $ 143+10=153 $. However, that does not give a remainder of $ 5 $ when divided by $ 7 $, so we add more $ 143 $s until we get a value that works. We get that $ 153+143+143=439 $ gives a remainder of $ 5 $ when divided by $ 7 $. Since we want the largest integer less than 2010, we keep adding the least common multiple of $ 7 $, $ 11 $, and $ 13 $ until we go over. The least common multiple is $ 7 \cdot 11 \cdot 13 =1001 $. We add it to $ 439 $ to get $ 1440 $, adding it again would give a value greater than $ 2010 $, so our answer is $ \boxed{1440} $.