Expression for a linear operator Let $\mathcal{B}=\{b_1, b_2, b_3\}$ be a basis of $\mathbb{R}^3$, 
where $b_1=(1,0,0); b_2=(1,1,0); b_3=(0,1,1)$.
Let $T:\mathbb{R}^3\to\mathbb{R}^3$ a linear operator with matrix (respect to the basis $\mathcal{B}$) is given by
$$A=\left[\begin{array}{ccc} 3 & -4 & 2 \\ 1 & -2 & 2 \\ 1 & -5 & 5 \end{array}\right].$$
and let $x,y,z$ the coordinates of a vector in $\mathbb{R}^3$ with respect to the basis $\mathcal{B}$.
The exercise asks to find the expression for $T(x,y,z)$.
Since the matrix is the matrix of $T$ with respect to $\mathcal{B}$ and since the coordinates are with respect to $\mathcal {B}$ to compute $T(x,y,z)$ is right say that
$T(x,y,z)=A(x,y,z)^T$?
 A: That depends on whether you take $(a,b,c)$ to mean $ae_1+be_2+ce_3$ or $ae_1'+be_2'+ce_3'$, where the primed vectors are the canonical versors of $\mathbb{R}^3$, that is $(1,0,0),(0,1,0),(0,0,1)$. I personally had overlooked this, but to be sure I would write that $T(x,y,z)$ is $A(x,y,z)^T$ if by $(x,z,y)$ we mean a vector of coordinates with respect to $\mathcal{B}$, and then compute the expression $A(x,y,z)^T$, which is (as you noted in a comment) $(3x-4y+2z,x-2y+2z,x-5y+5z)$. Without specifying that vectors of coordinates are all given w.r.t. $\mathcal{B}$, someone very precise might complain that coordinate vectors to him/her are only in the canonical basis.
In fact, for that person $T(x,y,z)$ might mean "take map $T$ and apply it to $(x,y,z)$ where the coordinates are with respect to the canonical basis". Which would require computing a change of basis and changing the basis for $A$. BUt it seems very unlikely. After all, the canonical basis is never mentioned, so I guess it is safe enough just to assume coordinate vectors are always w.r.t. $\mathcal{B}$ in this exercise and its solution.
Bottom line: either you did right, or whoever says otherwise is being absurdly pernickety if you ask me :).
