Find equation of plane given plane point

Question

So I am given one plane point $M(5,2,0)$, also two points which are not plane points: $P(6,1,-1)$ distance to plane $1$, also point $Q(0,5,4)$ distance to plane $3$. How find equation of plane with the information given?

Answer 1

Hint.

Suppose that an equation of the plane $\mathscr P$ to be find is $\mathscr P \equiv ax+by+cz+d=0$ where $a^2+b^2+c^2=1$. By hypothesis, $M$ belongs to the plane. Hence $5a+2b+d=0$.

Also the distance of a point of coordinates $(x,y,z)$ to $\mathscr P$ is $\vert ax+by+cz+d \vert$. Therefore we also have $$\begin{cases} \vert 6a+b-c+d \vert = 1\\ \vert 5b +4c+d \vert =3 \end{cases}$$ Replacing with $d=-5a-2b$ we get $$\begin{cases} \vert a-b-c\vert = 1\\ \vert 5a-3b -4c\vert =3 \end{cases}$$ So you have to solve the following system of equations: $$\begin{cases} a^2+b^2+c^2=1\\ \vert a-b-c\vert = 1\\ \vert 5a-3b -4c\vert =3 \end{cases}$$