Please, can you check the following execution is correct:
Problem text
I have a plane in affine space in $\Bbb R^4$ described by two following equations:
\begin{Bmatrix}3x+y-z-q +1=0\\ -x-y+z+2q=0\end{Bmatrix}
I have the coords of a point $P$: (0,1,1,0) Describe the locus of $Q$ points such that line $PQ$ is $\bot$ to the plane.
My solution
Now I look for the line (which $Q$ points belong to) perpendicular to the plane and passing trough point $P$. In order for a plane described by equation $ax+by+cz+dq+e=0$ and a line whose coefficients are $l,m,n,t$ to be perpendicular, this must be true:
$\frac{a}{l}=\frac{b}{m}=\frac{c}{n}=\frac{d}{t}$
So, the equations of line passing by generic point $P(x_0,y_0,z_0,q_0)$ is:
$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}=\frac{q-q_0}{d}$
Then, in my specific problem I get:
$\begin{Bmatrix}x=y-1 \\ y=2-z \\ z=\frac{q}{2}+1 \end{Bmatrix}$