Find a plane perpendicular to a plane passing by point In $\mathbb R^4$ I have: $$\pi: \begin{cases} x+y-z+q+1=0 \\ 2x+3y+z-3q=0\end{cases}$$
I have to find $\pi' \bot$ $ \pi $ and passing by $P=(0,1,0,1)$. How can I do that? Thanks a lot!

The equation of any plane$(\pi_1)$ passing through $P(0,1,0,1)$ is $a(x-0)+b(y-1)+c(z-0)+d(q-1)=0$ where $a,b,c,d$ are indeterminate constants,
If  $\pi_1 \bot \pi $ , the sum of the product of the directional cosines will be $0$.
So, $(1)(a)+(1)(b)+(-1)(c)+(1)(d)=0\implies a+b-c+d=0$
and $(2)(a)+(3)(b)+(1)(c)+(-3)(d)=0\implies 2a+3b+c-3d=0 $
So, $2c=5a+6b$  and $2d=3a+4b$,
So, $\pi_1$ becomes $2a(x-0)+2b(y-1)+(5a+6b)(z-0)+(3a+4b)(q-1)=0$
Or, $a(2x+5z+3q-3)+b(2y+6z-2+4q-4)=0$
Or, $a(2x+5z+3q-3)+b(2y+6z+4q-6)=0$
If $ab≠0$, the equation of the plane $\pi_1$  will be $2x+5z+3q-3=0$ and $2y+6z+4q-6=0.$
