Shell Programming and Scripting

Hello im new here and i shot stright with question.
Mainly i wanna ask , how do i search with regexp in one spot and show the whole thing, what im trying to ask is , for eg. i do ls -l, and i see all the info for the dirs and dats. now say i wanna get all the dats that in their name they start with number and end with extension .pt and on all of that they need to have rights for execution for all i mean group users, user and admin.. so i wanned to know is there way that i can firstly separate all the dats with the rights for ex. like grep '^-..x..x..x.*' and than in the filtered income to look only by the name but show the whole line that ls -l gives, i know i can use cut and take just the name column but this way i lose all the rest of the output i just get the name , i dont like that , is there way to get the power of cut but not delete the other content of the line like looking by the name column and if it meets the criteria i require than print the whole line that ls -l gave.. thanks!

Try:

Code:

find . -type f -name "[0-9]*.pt" -perm -111 | xargs ls -l

Code:

ls -l [0-9]*.pt |grep '^-..x..x..x'