Consider a linear operator $T:\mathbb{R^3}\rightarrow\mathbb{R^3}$ with a standard matrix
$$ A = \left[\begin{array}{rrr} 1 & 2 & 3\\ 2 & 1 & 3\\ 1 & 3 & 2\\ \end{array}\right] $$
Let ${e_1, e_2, e_3}$ denote the standard basis for $\mathbb{R^3}$.
a) Find $T(e_j)$ for every $j=1,2,3$.
First, your confusing the rows of A as being the values of the entries of $T(e_{j})$. So it is incorrect to say that $T(e_{1})=(1,2,3),T(e_{2})=(2,1,3),T(e_{3})=(1,3,2)$.
Recall that $T(e_{j})$ is a vector, which is an ordered pair of entries, and in your problem, the matrix $\,A$ is the $3 \times 3$ matrix whose $j$-th column is the vector $T(e_{j})$ for $j=1,2,3$. So $$A=\begin{bmatrix}T(e_{1}) & T(e_{2}) & T(e_{3})\end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 1 & 3 & 2 \end{bmatrix}$$
and by inspection we see that $$T(e_{1})=\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}=(1,2,1), T(e_{2})=\begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}=(2,1,3), T(e_{3})=\begin{bmatrix} 3 \\ 3 \\ 2 \end{bmatrix}= (3,3,2).$$
b) Find $T(2e_1+ 5e_2+ 3e_3)$.
In here, recall that the vectors $e_{1},e_{2},e_{3}$ corresponds to the columns of the $3\times 3$ identity matrix, so $e_{1}=(1,0,0),e_{2}=(0,1,0),e_{3}=(0,0,1)$. Your mistake here is in assigning these columns of the identity matrix with the values of the rows of $A$ and in multiplying that with the given weights of $e_{1},e_{2},e_{3}$.
Instead, what we need to do is to use our knowledge of $T(e_{1}),T(e_{2}),T(e_{3})$ to determine $T(x)$ given any vector $x$. In general, for a $3\times 3$ matrix, we have $$T(x)=\begin{bmatrix}T(e_{1})& T(e_{2}) &T(e_{3})\end{bmatrix}\begin{bmatrix} x_{1}\\x_{2}\\x_{3}\end{bmatrix}=T(x_{1}e_{1}+x_{2}e_{2}+x_{3}e_{3}).$$
So using what we know from (a) and the linearity of $T$:
$$\begin{align}T(2,5,3)&=T(2e_1+ 5e_2+ 3e_3)\\ &=2T(e_{1})+5T(e_{2})+3T(e_{3})\\ &=2(1,2,1)+5(2,1,3)+3(3,3,2)\\ &=(2,4,2)+(10,5,15)+(9,9,6)\\ &=(21,18,23)\\ \end{align}$$
