Since [tex]T[/tex] is linear, we know that [tex]T\left(\vec{u} + \vec{v}\right) = T\left(\vec{u}\right) + T\left(\vec{v}\right)[/tex].
Since [tex]T[/tex] is linear, we know that [tex]T\left(c \vec{v}\right) = c T\left(\vec{v}\right)[/tex], for any real scalar c.
You can find the standard vectors as linear combinations of the given vectors by constructing an augmented matrix and row reducing, as you did.
For example:
[tex]\begin{pmatrix}1&& -1&& 1&& 1&&\\2&& -4&& 5&& 0&&\\ 2&& -5&& 7&& 0&&\end{pmatrix} -> \begin{pmatrix}1&& 0&& 0&& 3&&\\0&& 1&& 0&& 4&&\\ 0&& 0&& 1&& 2&&\end{pmatrix}[/tex]
So we can write [tex]\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}[/tex] as a linear combination of [tex](3)\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} + (4)\begin{bmatrix} -1 \\ -4 \\ -5 \end{bmatrix} + (2)\begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}[/tex]
Now we know that:
[tex]T\left(\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}\right) = T\left( (3)\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} + (4)\begin{bmatrix}-1 \\ -4 \\ -5 \end{bmatrix} + (2)\begin{bmatrix}1 \\ 5 \\ 7 \end{bmatrix}\right) = (3)T\left( \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} \right) + (2)T\left(\begin{bmatrix}-1 \\ -4 \\ -5 \end{bmatrix} \right) + (4)T\left(\begin{bmatrix}1 \\ 5 \\ 7 \end{bmatrix} \right)[/tex]
So, what is:
[tex](3)T\left( \begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} \right) + (4)T\left(\begin{bmatrix}-1 \\ -4 \\ -5 \end{bmatrix} \right) + (2)T\left(\begin{bmatrix}1 \\ 5 \\ 7 \end{bmatrix} \right)[/tex]
The values for the other standard vectors can be found with a similar process.