Systematically finding a basis of a subspace Learning Linear Algebra on my own. Going through MIT Open Courseware lectures. There is a problem with solution and I understand the answer but would like to find a more systematic way to find it.
Find 


*

*a basis for the plane $x − 2y + 3z = 0$ in $R^3$

*a basis for the intersection of that plane with the $xy$ plane


Solution 
Part a
This plane is the nullspace of the matrix $\begin{bmatrix}1&-2& 3\end{bmatrix}$ and also $A=\begin{bmatrix}1 & −2 & 3 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$
The special solutions to $Ax = 0$ are
$v_1 = \begin{bmatrix}2 \\ 1 \\ 0 \end{bmatrix}$ and $v_2 = \begin{bmatrix} -3\\ 0 \\ 1\end{bmatrix}$
These form a basis for the nullspace of $A$ and thus for the plane. 
Part b
The intersection of this plane with the $xy$ plane contains $v_1$ and does
not contain $v_2$; the intersection must be a line. Since $v_1$ lies on this line it also provides a basis for it. 
I understand how both parts were solved, but my question is about part b. Is there a more systematic way to find a solution? Say I can't see immediately that $v_1$ is in the plane. How would I solve the problem?
I did try solving the following system
$\left \{\begin{array}{l}x-2y+3z=0\\x+y=0\end{array}\right.$
Wrote it in matrix form
$
\begin{bmatrix}
1 & -2 & 3 \\
1 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix} 
$
And then reduced it to the row reduced echelon form and got the following
$
\begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{bmatrix} 
$
Which led me to believe that the answer should be 
$
\begin{bmatrix}
1 \\
-1 \\
-1
\end{bmatrix} 
$
But clearly I'm wrong. It seems to me I'm missing some important concept. I would really appreciate any help on this. 
 A: You correctly applied the definition of the "$xy$-plane" in your first approach, but not in the second.  A point $(x,y,z)$ is on the $xy$-plane if and only if $z = 0$.  As such, our system of equations should have been
$$
\left \{\begin{array}{l}x-2y+3z=0\\z=0\end{array}\right.
$$
Corresponding to the system of equations $Ax = 0$ with 
$$
A = \pmatrix{1 & -2 & 3\\0 & 0 & 1}
$$
(by the way, I really don't understand your apparent preference for square matrices).  From there, find the row-echelon form and confirm that $v_1 = (2,1,0)^T$ is our only special solution, so this vector forms a basis of the desired nullspace.
A: Perhaps it is just because I learned Linear Algebra as being about vector spaces and learned matrices as a specific notation, but I have always disliked writing things in terms of matrices.  Here is how I would do exercise (1):
The equation x- 2y+ 3z= 0 can be written x= 2y- 3z so any point in that plane can be written (x, y, z)= (2y- 3z, y, z)= (2y, y, 0)+ (-3z, 0, z)= y(2, 1, 0)+ z(-3, 0, 1).  So a basis for that plane is {(2, 1, 0), (-3, 0, 1)}.
Now, the intersection of that with x+ y= 0 or x= -y, allows us to write x- 2y+ 3z= 0 as -y- 2y+ 3z= 0 or -3y+ 3z= 0 or, finally, z= y.  Then y(2, 1, 0)+ z(-3, 0, 1) becomes y(2, 1, 0)+ y(-3, 0, 1)= y(-1, 1, 1).  That is, that one-dimensional space has {(-1, 1, 1)} as basis.
A: A basis for the plane $x − 2y + 3z = 0$ consists of two vectors perpendicular to $(1,-2,3).$ Thus we have
$$\{(0,3,2),(3,0,-1)\}, \:\: \{(0,3,2),(2,1,0)\}, \:\: \{(3,0,-1),(2,1,0)\}$$ as possible basis.
A basis for the intersection of planes $$\begin{cases}x-2y+3z&=0\\z&=0\end{cases}$$ consists of one vector satisfying both equations. Thus $$\{(2,1,0)\}$$ is a basis.
