I'm trying to get a system of equations that defines the following line $r$. Here is its parametric equation
$r: (1,2,3)+t(3,1,2) \mid t \in \mathbb{R}$
To find the two equations I calculate the nullspace of $(3,1,2)$, this gives me the coefficients of the equations.
(nullspace) $\mathscr{N}(3,1,2)=\mathscr{L}((-2,0,3),(-1,3,0))$
Therefore the two equations should be
$r: \begin{cases} -2(x-1)+3(z-3)=0\\ -(x-1)+3(y-2)=0 \end{cases}$
But if I solve the system I get: $\mathscr{L} ((1,\frac{1}{3},\frac{2}{3})) + (0,\frac{5}{3},\frac{7}{3})=(0,\frac{5}{3},\frac{7}{3})+ t (1,\frac{1}{3},\frac{2}{3}) \mid t \in \mathbb{R}$
Which is not the original equation of $r$.
Where is my mistake?
Thanks a lot in advice
What you get is a different representation of the same line.
Compare
\begin{align*} (1,2,3)+t(3,1,2) &= (0,\tfrac53,\tfrac73)+u(1,\tfrac13,\tfrac23) \end{align*}
From the first coordinate you can read that $u=1+3t$. If you plug that into the other coordinates, you will find that the equation holds in this case as well. So it's just a different representation of the same line.