Consider a linear transformation $T:\mathbb{R}^3\to\mathbb{R}^3$ satisfying $$T \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} \ \text{and} \ T \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 5 \\ 6 \\ 7 \end{pmatrix}. $$
The question told me to find $T\begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix}$ which I did but I am wondering how would I find the transformation matrix $T$?
Hint:It amounts to finding $a,b$ such that: $a\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + b\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix}$. Or:
$\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$. This takes care of the question being asked.
To find $T$, we have to find: $T(e_1), T(e_2), T(e_3)$ by similar way. For example:
Find $x,y,z$ such that: $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = x\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + y\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix} + z\begin{pmatrix} 3 \\ 4 \\ 5\end{pmatrix}$
Note: As Rebecca J. Stones pointed out that you have insufficient information to determine $T$. You need to know a value of $T$ at another vector (that is linearly independent with the first two) to complete the solution.
Linearity give us $$T\begin{pmatrix}1\\1\\1\end{pmatrix}=T\begin{pmatrix}2\\3\\4\end{pmatrix}-T\begin{pmatrix}1\\2\\3\end{pmatrix}=\begin{pmatrix}5\\6\\7\end{pmatrix}-\begin{pmatrix}4\\5\\6\end{pmatrix}=\begin{pmatrix}1\\1\\1\end{pmatrix}$$ Then $$T\begin{pmatrix}3\\4\\5\end{pmatrix}=T\begin{pmatrix}2\\3\\4\end{pmatrix}+T\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}5\\6\\7\end{pmatrix}+\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}6\\7\\8\end{pmatrix}$$ In order to determine $T$ completely we need the image of some basis $\mathcal{B}$ of $\mathbb{R}^3$ under $T$.
There is insufficient information to uniquely determine the transformation matrix. The transformation matrix has $9$ unknowns and we are given only $6$ linear equations.
It's equivalent to solving the underdetermined system of linear equations
$$ \left[ \begin{array}{ccccccccc|c} 1 & 2 & 3 & 0 & 0 & 0 & 0 & 0 & 0 & 4 \\ 0 & 0 & 0 & 1 & 2 & 3 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 2 & 3 & 6 \\ 2 & 3 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 5 \\ 0 & 0 & 0 & 2 & 3 & 4 & 0 & 0 & 0 & 6 \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 3 & 4 & 7 \\ \end{array} \right] $$
If we were to attempt to find it (and it would be tedious to do by hand), we would get a parameterized transformation matrix with $3$ parameters.
It's also not possible to uniquely find e.g. $T(e_1)$, since $$\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix},\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}, \text{ and } \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$ are linearly independent, i.e., there are no solutions to $$a\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}+b\begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}.$$
