Find a plane that contains two points and parallel to a line Find All planes that contain the two points P(2,-1,3), Q(1,0,1) and is parallel to the line x(t)=(2,2,-3)+t(-1,1,-2).
I thought I'd approach it by getting the vector PQ, but PQ is the same vector as the one given in the parallel line. So because the two points and the line are also parallel, this means that the plane can basically only rotate around the two points right? Geometrically, I thought that made sense. I just can't figure out how to get a formula. Help is appreciated.
 A: The line does not tell us anything. So you are right, the two points $(2,-1,3)$ and $(1,0,1)$ being on the plane are the only conditions.
This means
$$A(x-2)+B(y+1)+C(z-3)=0$$
$$Ax+By+Cz=2A-B+3C$$
And
$$A(x-1)+By+C(z-1)=0$$
$$Ax+By+Cz=A+C$$
Combine these two equations to see what you get.
$$A+C=2A-B+3C$$
$$-A=-B+2C$$
$$A=B-2C$$
So the planes are of the form,
$$(B-2C)x+By+Cz=B-C$$
With $B,C \in \mathbb{R}$.
A: This solution follows the method of 'user' on https://tinyurl.com/twddfdvx
Line p: (2,-1,3)+t(-1,1,-2) is in the plane
Thus :-a+b-2c = 0 and b=a+2c
Using the equation of the plane: ax+(a+2c)y+cz+d = 0 containing (2,-1,3)
2a+(a+2c)(-1)+c(3)+d=0 which gives d=-a-c
ax+(a+2c)y+cz-a-c = 0
Set a=1, c=k
x+(1+2k)y+kz-1-k=0.....(A)
For mutually perpendicular, dot product is zero:
(-1,1,-2).(1,1+2k,k) gives k=0
On putting k=0 in equation 'A' above gives x+y-1 = 0
This equation satisfies the points in the plane (2,-1,3) and (1,0,1)
The line in the plane and the parallel line are both parallel so cross product calculation can't be used. Is there another method?
