Error --> could not find installable ISAM --> TRANSFERDATABASE (ODBC)

I am trying to programmatically link SQL SERVER table to a MS ACCESS Database.
Here is my code -->

Function linktable()
        Dim oAccess As Access.Application
        oAccess = New Access.ApplicationClass

       ' Open a database:
        oAccess.OpenCurrentDatabase(filepath:="C:\Documents and Settings\abdullah\Desktop\dbtest.mdb", Exclusive:=False)

        Dim constr
        constr = "ODBC, Description=qqq; Driver = SQL Server; Server=Abdullah; APP = Microsoft Office XP; WSID = ABDULLAH; DATABASE=VSKSUS; TRUSTED_CONNECTION = YES"

        oAccess.DoCmd.TransferDatabase _
        (Access.AcDataTransferType.acLink, _
         "ODBC Database", _
          Access.AcObjectType.acTable, _
          "CustomerInfo", _
          "CustomerInfo", _
          True, True)

End Function

--> I get the error on the DoCmd link sayin "could not find installable ISAM"
I have already checked the Microsoft Support page on the matter -->

But all the registry keys seem to be fine
Does anybody have a solution or a way around this?

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iboutchkineConnect With a Mentor Commented:

The DLL for an installable ISAM file could not be found. This file is required for linking external tables
(other than ODBC or Microsoft Jet database tables). The locations for all ISAM drivers are maintained in the
Windows Registry. These entries are created automatically when you install your application. If you change
the location of these drivers, you need to correct your application Setup program to reflect this change and
make the correct entries in the Registry.

Possible causes:

An entry in the Registry is not valid. For example, this error occurs if you're using a Paradox external
database and the Paradox entry points to a nonexistent directory or driver. Exit the application, correct the
Windows Registry, and try the operation again.

One of the entries in the Registry points to a network drive and that network is not connected. Make sure
the network is available, and then try the operation again.

Another cause
The error occures when you are using a different version ACCESS than the original code did so the
"connect" property of the data-control is set wrong. Just set it ot the correct one

also check
Bob LearnedConnect With a Mentor Commented:
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