3D Equations of Planes 
So the above question is from a Calc III class I'm taking but I'm not sure I understand the solution...
$n_1$ represents the vector normal to the given plane. Therefore taking the cross product of $n_1$ and the vector $P_1P_2$ would output a vector that is parallel to the given plane, not orthogonal as desired. Unless I'm mistaken...
Anyone care to elaborate? Are both solutions correct?
Here was my attempted solution:
Let $r$ be the vector $<3,-1,-1>$ and let $M_1$ and $M_2$ be arbitrary points satisfying the equation of the plane given. I chose $M_1 = (3,10,0)$ and $M_2 = (2,6,0)$ therefore let $m$ be the corresponding vector $m = <-1,-4,0>$ then $$r \times m = <-4,1,-13>$$
which gives a final equation of $$ -4(x+2) + y-1 -13(z-4)$$
 A: We need to find the plane which is perpendicular to the plane $4x-y+3z=2$ and passing through the points $P_1(-2,1,4)$ and $P_2(1,0,3)$.
The plane can be written as $n\cdot(r-p)=0$ where $n$ is orthogonal to the plane and $p$ is a point on the plane. Given that we know $P_1$ and $P_2$ lie on the plane, we can choose $p$ as either of these points. Arbitrarily choose $p=P_2=(1,0,3)$ (it is just as correct to choose $p=P_1$). We also have $r=(x,y,z)$ as an arbitrary point on the plane.
The plane $4x-y+3z=2$ can also be expressed as $(4,-1,3)\cdot(r-q)=0$ for some point $q$ on the plane. Hence, a normal to the first plane is $n_o=(4,-1,3)$. The requirement now is that $n$ is orthogonal to $n_o$, since $n_o$ is parallel to the given plane, which is orthogonal to the required plane.
Also, since $P_1$ and $P_2$ lie in the plane, the vector joining the two points is parallel to the required plane. Hence, the vector $a=(3,-1,-1)$ is also parallel to the required plane.
This leads to $n$ needing to be orthogonal to $n_o$ and $a$. Thus $n=n_0\times a=(4,13,-1)$.
The required plane, is then $4(x-1)+13(y-0)-(z-3)=0$ or $4x+13y-z=1$.
