Is T onto? Explain (attempt inside)

Let T:R^2 -> R^3 be defined by T(a,b) = (a+b, a+2b, 2a + 3b)

is T onto?

So I first found out that T(1,1) would be, which came out to be (2,3,5)
so (2,3,5) is in the range of T. Then I asked myself if (2,4,6) would be in the range of T.

so I put it in matrix form after I equated it to (2,4,6)
a + b =2
a + 2b=4
2a+3b=6

after I put it in matrix form and reduced it, I got
1st row = (1 0 0 )
2nd row = (0 1 2)
3rd row= (0 0 0 )

Does this mean that it's not onto because there are only 2 non-zero rows implying that it's not in R^3?

I'm confused as to see why this is not onto (answer is in the back of the book which says it's not onto)

Any help would be appreciated. Thanks.

First we see how the standard unit vectors transform under the transformation T.

T(e1)=(1,1,2)
T(e2)=(1,2,3)

Therefore the matrix representation of T wrt the standard basis is:

A=
[1 1]
[1 2]
[2 3]

By row reducing this matrix we see that

rref(A)=
[1 0]
[0 1]
[0 0]

So we can see that rankA=2 and therefore dimR(A) (dimension of the column space of A) is rankA = 2. dimR^3 = 3 < 2 which implies that R(A) is a strict subset of R^3 (ie does not equal R^3) and therefore the columns of A do not span R^3

Therefore T is not onto //