I'm reviewing for the math GRE. Here's a linear algebra question from an old text.
"Find an equation for the plane that is perpendicular to the plane $$8x-2y+6z=1$$ and passes through the points $P_1(-1,2,5)$ and $P_2(2,1,4).$"
Since our plane is perpendicular to a plane with normal vector $\langle8,-2,6\rangle,$ our plane's normal vector is a scalar multiple of the vector $\langle 1,1,-1\rangle$ (since $\langle 8,-2,6\rangle \cdot \langle1,1,-1\rangle=0$).
Since we have a normal vector and a point $P_1$, our plane is uniquely determined: Given any point $(x,y,z)$ on our plane, the equation of our plane is $$\langle 1,1,-1\rangle \cdot\langle x+1,y-2,z-5\rangle =0$$
Simplifying we get $$x+y-z+4=0$$
However, this is the wrong plane since $P_2$ is not on this plane. Where did I go wrong?