How do I open a file using process.start?

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VB

I have some code that can open a file with either Notepad or Wordpad. I am attempting to be able to open the same file with a text editor of the users choice. I tried Microsoft Word as a test but cannot get it to work. Here is my code (oh: the variable "OtherPuzzleEditorPath" is defined as Public in a module):

Private Sub viewBtn_Click(sender As Object, e As EventArgs) Handles viewBtn.Click
        Dim MySelectedPath As String = Nothing
        Dim MySelectedExt As String = Nothing
        Dim EditorPath As String = Nothing
        
        Select Case True

            Case RadioButton1.Checked
                MySelectedPath = PuzzlePath
                MySelectedExt = ".wsp"

            Case RadioButton2.Checked
                MySelectedPath = SolutionsPath
                MySelectedExt = ".wss"

        End Select

        EditorPath = MySelectedPath & ListBox3.SelectedItem.ToString & MySelectedExt

        Select Case True

            Case OpenWithNotepad
                Process.Start("Notepad.exe", EditorPath)

            Case OpenWithWordpad
                Process.Start("Wordpad.exe", Chr(34) & EditorPath & Chr(34)) 'DONT CHANGE THIS LINE!

            Case OpenWithOther
                Process.Start(OtherPuzzleEditorPath & Space(1) & EditorPath)

        End Select

        ListBox3.ClearSelected()
        deleteBtn.Enabled = False
        viewBtn.Enabled = False

    End Sub<pre lang="vb">

What I have tried:

Different variations of defining both strings: ie(Chr(34) & [path] & Chr(34)), changing to the directory the executable is in FIRST, then running it with the path of the file.