In $r_1\Join r_2$ $ C$ is common ,in which C is foreign key ( assuming that foreign key is present) in r1 so in natural join of r1 and r2 all record of r1 will be present so $r_1\Join r_2$ has maximum no. of tupples 150 .let say $r_1\Join r_2$ as S ,now join of S and r3, in which no element is common( assuming all attributes have diffrent quality as in question nothing is mentioned) so $S\Join r_3$ will become $S \textbf{*}r_3$(cross product) .now every tupple of S will relate to every tupple of $r_3$ ,so maximum no. of tupples =150*75=11250.