If you get all three codes together you have 15 numbers and the rest is all letters. 15 numbers is exactly what you need for 3 passcodes.

ekh9clutc78xqj6iraro38dsg22myre84uub99appi777yrh

A 12×4 grid reveals interesting things

In the first column you can see a good start for a passcode. Then in other columns you can see “cryp” which could lead to the keyword crypto. Let’s rearrange the columns as if we were using a skip of 3: 1, 4, 7, 10, 2, 5, 8, 11, 3, 6, 9, 12

Read going down from the upper left E

eqg99impure778uykj29crypto878durh62alaric347xsbh

Still need a bit of tweaking but let’s split the string in 3 equal parts

eqg99impure778uy
kj29crypto878dur
h62alaric347xsbh

The first row is a valid passcode, for the 2nd row move the r at the begining and for the 3rd row move bh to the begining

eqg99impure778uy

rkj29crypto878du

bhh62alaric347xs