ok, we have two sets:
the first task is to show that B' is indeed a basis:
since we know , it suffices to show that B' is linearly independent.
so suppose , that is:
hence , so thus and thus .
so B' is indeed linearly independent and thus a basis for .
now if T is the matrix that sends (so that ),
then clearly the matrix that turns B-coordinates into B'-coordinates is .
you indeed have found T correctly, and so now we must find T-1.
we COULD use row-reduction to do this, but an easier way is to express the as linear combinations of the .
now (1,0,0) = (1/2)(1,0,1) + (-1/2)(0,1,1) + (1/2)(1,1,0)
(0,1,0) = (-1/2)(1,0,1) + (1/2)(0,1,1) + (1/2)(1,1,0) and
(0,0,1) = (1/2)(1,0,1) + (1/2)(0,1,1) + (-1/2)(1,1,0), so:
this agrees with what you found.
now B-coordinates are just "the usual cooordinates" that is: .
so the B-coordinates of are: (1,0,0) + 2(1,1,0) = (1,0,0) + (2,2,0) = (3,2,0).
however, the B'-coordinates of will be:
so the answer you were given is correct (a sign error perhaps?)