Linear Transformation from $\alpha$ to $\beta$ T: $R^3$ $\to$ $R^2$
$$[T]_{\beta\alpha} = 
        \begin{matrix}
        2 & 3 & 1 \\
        1 & 2 & 1 \\
        \end{matrix}
$$
$\alpha$ = {(1, -1, 1), (0, 1, 0), (1, 0, 0)}
$\beta$ = {(3, 2), (2, 1)}
Find: T((x, y, z))   for any x, y, z in $R^3$
My approach:

$[T(v)]_{\beta}$ = $[T]_{\beta\alpha}$$[v]_{\alpha}$

so,

$ z\begin{pmatrix}
        1 \\
        -1 \\
          1       \\
        \end{pmatrix}  +    
(y+z)\begin{pmatrix}
        0 \\
        1 \\
        0       \\
        \end{pmatrix}
+(x-z)\begin{pmatrix}
        1 \\
        0 \\
         0       \\
        \end{pmatrix} = \begin{pmatrix}
        x \\
        y \\
        z       \\
        \end{pmatrix}  $

and

$[(x, y, z)]_{\alpha} = [(z), (y + z), (x - z)]$

then

$[T]_{\beta\alpha}[\begin{pmatrix}x\\y\\z\\\end{pmatrix}]_{\alpha}=\begin{pmatrix}x + 3y + 4z\\x + 2y + 2z\\\end{pmatrix}$

finally, 

$(x + 3y + 4z)\begin{pmatrix}3\\2\\\end{pmatrix} + (x + 2y + 2z)\begin{pmatrix}2\\1\\\end{pmatrix} = \begin{pmatrix}5x + 13y + 16z\\3x + 8y + 10z\\\end{pmatrix}$

and

T((x, y, z)) = ((5x + 13y + 16z), (3x + 8y + 10z))

But this does not look like it makes lot of sense, I think I misunderstand something.
Can anyone give me a pointer?

it is correct in my humble opinion except rhe last step.
finally, 
$$(x + 3y + 4z)\begin{pmatrix}2\\1\\\end{pmatrix} + (x + 2y + 2z)\begin{pmatrix}3\\2\\\end{pmatrix} =...$$
instead of 
$$(x + 3y + 4z)\begin{pmatrix}3\\2\\\end{pmatrix} + (x + 2y + 2z)\begin{pmatrix}2\\1\\\end{pmatrix} = \begin{pmatrix}5x + 13y + 16z\\3x + 8y + 10z\\\end{pmatrix}$$
