I think the way you have organized the table, each value corresponds to another one in the same row.
Maybe you only need one ListPicker.
When you select the first one, the second and third ones are in the same row.
I think I would use 3 Spreedsheet components, one for each column.
With each component you get each of the columns.
Element 1 in the first column corresponds to element 1 in the second column and element 1 in the third column.
If you select the 5th element in the first ListPicker, the 5th element in the second list corresponds to the value of the same row.
In short, each column you get is a list. If in list 1 you select object number 5, in the second and third list you must select element 5…
Each result of the second and third step can be shown on a label.