Three numbers a,b,c are in A.P. ad they are used to making 9− digit number using each digit thrice such that atleast three consecutive digits are in A.P., So, c,b,a will also be in A.P.
Now, we have 7 position to put (a,b,c)or(c,b,a) as \underset{7\mathrm{positions}}{\underset{⏟}{_______}}__.
Number of such numbers is
=C12⋅C17⋅2!⋅2!⋅2!6!
=1260