Your answer is correct, as you can check doing these calculations:
$ A \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} = A \begin{pmatrix} -3 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} $
and
$ A \begin{pmatrix} 1 / 2 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 10 \\ 15 \end{pmatrix} \ . $
As for the answer given by your professor, you must take into account that those vectors $(-2 ,0 , 1), (-3,0, 1)$ and $(1/2, 0 ,0)$ are not unique. What is unique is the solution set
$ (1/2, 0 ,0) + \ \mathrm{span}\left\{(-2 ,0 , 1), (-3,0, 1) \right\} \ . $
But there are infinitely many ways to write it. For instance, another way to write the same solution set could be the following:
$ (0, 0, 1/6) +\ \mathrm{span}\left\{(-1 ,0 , 1/2), (-1,0, 1/3) \right\} \ . $
EDIT. I forgot: in order to fully check that your solution is right, you should also verify that the rank of your matrix $A$ is one (as it is); so the solution set has indeed dimension = number of unknowns - $\mathrm{rank}\ A = 3 -1 = 2$.