Compute $\det{T}$ where $T(X)=AX+XA$

One approach here is to "guess" the eigenvectors:

If $u$ is an eigenvector of $A$ and $v$ an eigenvector of $A^T$, then the matrix $uv^T$ is an "eigenmatrix" of the transformation $T$.

Now that you know the eigenvectors, you can find the eigenvalues, and find that the determinant is the product of all eigenvalues.

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A more concise approach is as follows: plug in $$ X= \pmatrix{1\\0\\0}\pmatrix{0&0&1} = \pmatrix {0&0&1\\0&0&0\\0&0&0} $$ to find that $T(X)=0$, which means that $T$ is not invertible, which means that $\det(T)=0$.

\begin{align} T(X) &= AX + XA \\ &= \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \end{pmatrix} + \begin{pmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \\ x_{31} & x_{32} & x_{33} \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{pmatrix} \\ &= \begin{pmatrix} x_{11}+ x_{21} & x_{12}+ x_{22} & x_{13}+ x_{23} \\ 2 x_{21} & 2 x_{22} & 2 x_{23} \\ -x_{31} & -x_{32} & -x_{33} \end{pmatrix} + \begin{pmatrix} x_{11} & x_{11} + 2x_{12} & -x_{13} \\ x_{21} & x_{21} + 2x_{22} & -x_{23} \\ x_{31} & x_{31} + 2x_{32} & -x_{33} \end{pmatrix} \\ &= \begin{pmatrix} 2x_{11}+ x_{21} & x_{11} + 3x_{12}+ x_{22} & x_{23} \\ 3 x_{21} & x_{21} + 4 x_{22} & x_{23} \\ 0 & x_{31} + x_{32} & -2x_{33} \end{pmatrix} \\ &= \begin{pmatrix} 2 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 3 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 4 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -2 \end{pmatrix} \begin{pmatrix} x_{11} \\ x_{12} \\ x_{13} \\ x_{21} \\ x_{22} \\ x_{23} \\ x_{31} \\ x_{32} \\ x_{33} \\ \end{pmatrix} \end{align} Row $7$ is zero, so $\det(T) = 0$.

For $A \in M_3(\mathbb{R})$, set $T_A(X) = AX + XA$. Note that your $A$ is diagonalizable and so we can find an invertible $P$ such that $P^{-1}AP = \operatorname{diag}(1,2,-1)$ := D. Consider the map $S \colon V \rightarrow V$ given by $S(X) = P^{-1}XP$. Note that $S$ is invertible and $S^{-1}(X) = PXP^{-1}$. Now,

$$ (S^{-1} \circ T_A \circ S)(X) = S^{-1}(T(P^{-1}XP)) = S^{-1}(AP^{-1}XP + P^{-1}XPA) = P(AP^{-1}XP + P^{-1}XPA)P^{-1} = DX + XD = T_D(X). $$

Thus, $T_A$ is similar to $T_D$ and $\det(T_A) = \det(T_D)$ so you can reduce the problem to the case the matrix is diagonal.