Do the vectors (1,3,-2),(2,1,1) belong to R(T)?

Let T:IR^3-->IR^3 be defined by T(a,b,c) = (a+b, b-2c, a + 2c). Do the vectors (1,3,-2), (2,1,1) belong to R(T)?

Looking through my notes I found some clues to this question. I found that "RT is always span (Tv1,..,Tvn) where v1,..,vn is basis of domain V.

My first approach was to make a column vector (a) = (a+b)

(b) (b-2c)

(c) (a+2c)

and then remove the coefficients like this:

a (1) b(1) c(0) to get a basis. I don't know if I was on the right track.

(0) (1) (-2)

(1) (0) (2)

The second thing I tried was to compute two vectors for R(T) like this:

T(1,3,-2) = (4,7,-3)

T(2,1,1) = (3,-1,4)

And then get that R(T) = span {(4,7,-3), (3,-1,4)}.

Anyways, I didn't finish my answer since I am confused. Also sorry about my notation because I don't know what syntax I need to use to make a matrix on this forum.