First off, your method does and does not work, depending on what you mean. What if the vector (0, 0, 0, 1) is already in the span of the 3 vectors you started with? Then row-reduction will lead to a 0 column. Essentially, what you did was guess that (0, 0, 0, 1) was not already in the span, and then you checked to see if you were right. This will work exactly in the cases when (0, 0, 0, 1) was not already in the span of the vectors you started with. As lhf points out in the comments to the question, this method will work a lot of the time. In fact, it would work for this problem if you had guessed any of (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), or (0, 0, 0, 1) (checked with Sage). But, it won't always work.
It is well known that the cross product of two vectors (in 3-dimensions) gives a new vector that is orthogonal to both of the starting two vectors. Therefore, if the first two vectors are linearly independent, and the new third one is orthogonal to both of them, then the set of 3 vectors is definitely linearly independent. By the way, if the first two are not linearly independent, then the cross product will be the 0 vector.
There is a generalization to the cross product that can be applied here. If in $n$-dimensions, use $n-1$ vectors, and the result will be a vector orthogonal to the $n-1$ vectors. So, to answer your question, one way would be to find the determinant:
$$\begin{vmatrix} i & j & k & l \\
1 & 2 & 0 & 2 \\
1 & 1 & 1 & 0 \\
2 & 0 & 1 & 3\end{vmatrix}$$
Assuming your first three vectors are linearly independent, the result will be a 4 dimensional vector (in terms of coordinates i, j, k, l) that is orthogonal to the 3 starting vectors.
I got $8i - j - 7k - 5l$, assuming I didn't make any mistakes. And, I checked in Sage that the 4 vectors would be linearly independent.
Reference: I learned about this when I took a 4th semester of calculus in college, where we used Vector Calculus by Susan Jane Colley. It is introduced in the exercises for Section 1.6. One of the exercises is to prove that the new vector is orthogonal to the previous ones.
If I understand the question, you want a spanning set for the $xy$-plane in ${\mathbb R^3}$. A spanning set for the $xy$-plane is
$${\mathcal{B}}=\left\{\begin{bmatrix}a\\b\\0\end{bmatrix},\,\begin{bmatrix}c\\d\\0\end{bmatrix}\right\},
$$
where $ad-bc\neq 0$. Any choice of $a,\,b\,c$ and $d$ which satitfy the condition are a spanning set.
Some examples:
$$\begin{align}\mathcal{B}_1&=\left\{\begin{bmatrix}1\\0\\0\end{bmatrix}, \,\begin{bmatrix}0\\1\\0\end{bmatrix}\right\}&\text{[the standard basis]}\\
\mathcal{B}_2&=\left\{\begin{bmatrix}1\\1\\0\end{bmatrix}, \,\begin{bmatrix}1\\-1\\0\end{bmatrix}\right\}\\
\mathcal{B}_3&=\left\{\begin{bmatrix}1\\-4\\0\end{bmatrix}, \,\begin{bmatrix}2\\1\\0\end{bmatrix}\right\}
\end{align}
$$
Best Answer
As in your first example, you have two (not one) free variable. You let $x= s$, $y = t$ in the first example and $z = t$ follow from the equation. In this example you must take $x$ but you can choose between $y$ and $z$ to be free for you.
In the second example you can take any two variables as free. Let's take $x = s$, $y = t$. The equation $x-2y+5z = 0$ gives $$ z = -\frac 15 x + \frac 25y = \frac 15(2t -s) $$ So a basis is given by $(1,0, -\frac 15)^\top$, $(0,1,\frac 25)^\top$.