I've tried to solve this:
Suppose there is a linear system which S is the solution set of it, $\displaystyle 0\in \mathbb{R}$ therefore $\displaystyle \left ( 0,0,0 \right ) \in S=\left \{ \left ( a,a^2,b \right )|a,b\in \mathbb{R} \right \}$.
So, now I can tell that this is homogeneous systems.
I'll show the contradiction,
$\displaystyle \left ( x_1,x_2,x_3 \right )\in S $ so proof of no existence of  liner system over R-gif.gif the linear system is homogeneous, so scalar multiplication is also a solution to the system. $\displaystyle -1 \in \mathbb{R}$ and proof of no existence of  liner system over R-gif-1-.gif
But $\displaystyle \forall a \in \mathbb{R}\rightarrow a^2\neq -x_2$.
therfore proof of no existence of  liner system over R-gif-2-.gif, and $\displaystyle S$ is not the solution set.

What do you think about my answer?