First we show that if one of the marked numbers is $5$, then $5$ is the second largest of the three marked numbers. For this purpose, let us note that if $5$ is marked for one of the columns, then there is exactly one number less than or equal to $4$ in this column, and therefore, there must be a column which contains at least two numbers less than or equal to $4$. For such a column the second largest number in it must be less than or equal to $4$, so the number marked for that column must be less than or equal to $4$. By the same reasoning we conclude that there must be a column for which the marked number is greater than or equal to $6$. Thus $5$ is the second largest of the marked numbers.
From the consideration made above, it is enough to find the number of arrangements of nine numbers having $5$ as one of the marked numbers. There are $3$ ways of choosing a column to contain $5$, and for the chosen column containing $5$ there are $4^2$ ways of choosing other two numbers in that column, and there are $6$ ways of permuting the chosen three numbers in that column. Finally, for each arrangement of three numbers in that column, there are $6!$ ways of arranging the remaining six numbers into the other two columns. Therefore, the total number of the desired arrangements is
$$3 \cdot 4^2 \cdot 6 \cdot (6!) = 207360.$$