Finding a parametric solution via three specific solutions. The linear equation $$x - y + 2z = 3$$ has $[3, 0, 0], [0, 1, 2]$, and $[6, 1, -1]$ as specific solutions. 

The complete set of solutions corresponds to the set of points in the plane determined by the given equation. If we set $y = s$ and $z = t$, then a parametric solution is given by $[3 + s - 2t, s, t]$.

How do they found this parametric solution?

Thank you.
\begin{align}
&amp;\space x-y+2z = 3 \\
&amp;\Rightarrow x = 3 - 2t + s \\
&amp;\Rightarrow [x,y,z]=[3-2t+s,s,t]
\end{align}

Note that I used $y = s, z = t$ in the last step.
$$x-y+2z=3\implies x=3+y-2z$$

They then switched $y\iff s$, and  $z\iff t$