Locus perpendicular to a plane in $\mathcal{R}^4$ I have solved an exercise but I'm not sure to have solved it perfectly. Could you check it? It's very important for me.. 
In $\mathcal{R}^4$ I have a plane $\pi$ and a point P. I have to find the locus of Q points such that line PQ is perpendicular to $\pi$. 
$$\pi:\begin{cases} 3x+y-z-q+1=0\\ -x-y+z+2q=0 \end{cases}$$
$P=(0,1,1,0)$
$P$ doesn't belong to $\pi$ because it doesn't satisfy first equation. 
If $PQ$ line has to be $\perp$ to $\pi$, the vector of direction of $PQ$ must be perpendicular to spanners of the plane. 
Spanners of plane are:
$v_1=(1, -5, 0, -2)$ and $v_2=(0, 1, 1, 0)$
Then, $PQ \cdot v_1=0$ and $PQ\cdot v_2=0$ 
But I can write $PQ=OQ-OP$ 
and so I obtain $\begin{cases} (OQ-OP)\cdot v_1=0 \\(OQ-OP)\cdot v_2=0\end{cases}$
But inner product is a bilinear form and so I can also write:
$\begin{cases} OQ\cdot v_1 - OP\cdot v_1=0 \\OQ\cdot v_2-OP\cdot v_2=0\end{cases}$
I can calculate $OP\cdot v_1 $ and $OP\cdot v_2$ and, if $OQ=(x, y, z, q)$, I obtain:
$\begin{cases} (x, y, z, q) \cdot (1, -5, 0, -2)=(0,1,1,0) \cdot (1,-5,0,-2) \\ (x,y,z,q) \cdot (0,1,1,0)=(0,1,1,0,) \cdot (0,1,1,0) \end{cases}$
$\begin{cases}x-5y-2q=-5 \\
y+z=2 \end{cases}$
I have solved this equations set and I have obtained that $OQ= t(1, 1/5, -1/5, 0) + s(0,-2/5, 2/5, 1) + (0, 1, 1, 0)$ (with $t, s$ belonging to $\mathbb {R}$). 
And so I can say that my locus is given by $Q$ points having coordinates given by $t(1, 1/5, -1/5, 0) + s(0,-2/5, 2/5, 1) + (0, 1, 1, 0)$. 
My locus is given by vectors perpendicular to $\pi$ and so I can say that my locus is the orthogonal complement of $\pi$, so it is a plane.
Is it all correct? Please, signal me all things that you think are wrong. Thank you! 
EDIT: 
I can develope the exercise in another way saying that if $PQ$ has to be perpendicular to $\pi$, $PQ$ has to be perpendicular to any vectors of $\pi$ and so $PQ$ belongs to orthogonal complement of $\pi$, so PQ is given by:
$PQ=t(w_1)+s(w_2)$ where $w_1$ and $w_2$ are spanners of othogonal complement of $\pi$. Now I can obtain that $Q=P+t(w_1)+s(w_2)$. Is it correct? Thanks again

It all seems good.
But, you jumped one step: how did you find your $v_1$ and $v_2$ spanner vectors of $\pi$?
