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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?

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Yes, all what you have done is correct !

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    The question was: "Is it all correct?". My answer: "Yes, all what you have done is correct !" So, why does my "answer" not provide an answer to the question ?? – 2017-01-30
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    Sorry, its my bad. I upvote you for that.--:) – 2017-01-30
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Ja, it is correct! In fact:$$ \begin{cases} x+6y-3z=0\\2x+12y-6z=0 \end{cases} \text{ is equivalent to } \begin{cases} x+6y-3z=0\\ \frac{2}{2}x+\frac{12}{2}y-\frac{6}{2}z=0 \end{cases}$$$$\text{and }\begin{cases} x+6y-3z=0\\ \frac{2}{2}x+\frac{12}{2}y-\frac{6}{2}z=0 \end{cases} \text{ is equivalent to } \begin{cases} x+6y-3z=0 \end{cases} $$

then $x=-6y+3z$ with $y,z \in \Bbb{R} $. Therefore with $v \in \Bbb{R}^3$ we have $$\begin{align} v\in V &\leftrightarrow v=(-6y+3z,y,z)=y(-6,1,0)+z(3,0,1) \,\text{, with }y,z \in \Bbb{R} \\ &\leftrightarrow v \in \langle ((-6,1,0),(3,0,1)) \rangle \end{align}$$ In addition $((-6,1,0),(3,0,1))$ is $\Bbb{R}$-linear independent then $((-6,1,0),(3,0,1))$ is basis for $V$ also $\dim_\Bbb{R}(V)=2$