How to find equation of a plane that is passing through 2 points and normal to other plane?? The given points are $M(3,-1,2)$ and $M1(0,1,2)$ , plane A is passing though this 2 points. Plane B $2x-y+2z-1=0$ and its normal to A. How to we find the equation of plane A?  I have tried with cross product of vector $MM1$ and vector $b=(2,-1,2)$, but it didn't work.
 A: I think it must work, when you do the cross product of $MM_1$ with $b$ you get
$n=(-3,2,0)×(2,-1,2)=(4,6,-1)$.
This vector $n$ is a normal vector to plane A since the two vectors $b$ and $MM_1$ belong to plane $A$.
So the equation of plane A is $4x+6y-z+c=0$.
Substitute the coordinates of $M$(or $M_1$) in the equation to get $c$.
The equation is
$4x+6y-z=4$.
A: Perpendicular Planes
If you have a plane $P_1$ with equation $a_1 x + b_1 y + c_1 z = d_1$ and two points $p_1=(x_1, y_1, z_1)$ and $p_2=(x_2, y_2, z_2)$, then to find the plane $P_2$ with equation $a_2 x + b_2 y + c_2 z = d_2$ which is perpendicular to $P_1$ and contains both $p_1$ and $p_2$, you can use the following facts:


*

*The vector $v_1:= (a_1, b_1, c_1)$ is perpendicular to $P_1$

*The vector $v_2:= (a_2, b_2, c_2)$ is perpendicular to $P_2$

*The fact that $P_1$ is perpendicular to $P_2$ implies that $v_1$ is perpendicular to $v_2$. 

*The vector $v_3 := p_2 - p_1$ is parallel to $P_2$, and thus $v_3$ is perpendicular to $v_2$.   

*The fact that $v_2$ is perpendicular to both $v_1$ and $v_3$ implies that $v_2$ is a multiple of the cross product of $v_1$ and $v_3$.  You can safely set $v_2=v_1 \times v_3$.  (You should think about why this is true.) 

*You can use fact 5 to give you acceptable values for $a_2$, $b_2$, and $c_2.$

*To figure out $d_2$, use the fact that $p_1 \cdot v_2= d_2$ or $p_2 \cdot v_2= d_2$.

*Fact 6 gave you $a_2$, $b_2$, and $c_2$.  Fact 7 gave you $d_2$, so the problem is solved.

