Points $A$ and $B$ have coordinates $(−1,\:2,\:5)$ and $(2,\:−2,\:11)$ respectively. The plane $p$ passes through $B$ and is perpendicular to $AB$.
Find an equation of $p$, giving your answer in the form $ax+by+cz=d$.
I can use $AB = OB - OA$ to find the normal of the plane, $(3,\:-4,\:6)$ to form the equation $3x-4y+6z=d$. When point $B$ is inserted into the equation, I get the right value of $d$ ($80$). However, $A$ gives $d=19$. Is there something wrong here, or is this alright?
You are only told that $B$ lies on the plane. You are not told that $A$ lies on the plane. In fact $A$ cannot lie on the plane since $AB$ is perpendicular to the plane, and the plane passes through $B$.
Your first answer is correct, i.e. $d=80$.
Your second answer, i.e. $d=19$, is the equation of the plane passing through $A$ and perpendicular to $AB$. But that's not what you've been asked to find.