Find a basis for the given subspace: confirmation I would like to have a confirmation if possible!
Let $V$ be the subspace of $\mathbb{R}^3$ given by the solution of the system
\begin{equation}
\begin{cases} x+6y-3z=0\\2x+12y-6z=0 \end{cases}
\end{equation}
Find a basis for $V$.
By solving the homogeneous system I can find $\infty^2$ solutions (as the two row vectors are linearly dependent) of the kind $(3z-6y,y,z)$, which can be rewritten as $z(3,0,1)+y(-6,1,0)$. Hence a basis is $(3,0,1),(-6,1,0)$ and the subspace is two-dimensional. 
An example of solution is $(0,1,2)$, obtained by observing that for $y=,z=$ one has $2(3,0,1)+1(-6,1,0)=(0,1,2)$.
Is it all correct?

Yes, all what you have done is correct !
