I am asked the following problem:
Find the equation of the plane that has distance $1$ from line $X: (1,0,2) + \lambda (1,0,2)$ and contains points $P=(1,1,-1)$ and $Q = (2,1,1)$
Let's call the normal vector of the plane $\vec{n}$ so that $\vec{n} = (a,b,c)$
What I have so far is:
I) Since the distance from the plane and the line is $1$, using the point $(1,0,2)$ from the line and the normal vector of the plane,
$$ d_{istance} = 1 = \frac{\vert 1 a + 0 b + 2 c + d \vert }{\sqrt{a^2+b^2+c^2}}\\ 1 = \frac{\vert a+2c+d \vert }{\sqrt{a^2+b^2+c^2}} $$
II) Since the plane must be parallel to the line, the dot product between the normal vector and $(1,0,2)$ must be zero:
$$ (a,b,c) \cdot (1,0,2) = 0\\ a+2c = 0 $$
III) Since points $P$ and $Q$ are contained on the plane,
$$ a+b-c+d=0\\ 2a+b+c+d = 0 $$
I'm stuck on this part: there are more variables than equations to solve (If we combine the equations in III we get equation II). Is there something else that I'm missing?
Textbook's answer: $y-1=0$ and $6x-2y-3z-7=0$
Thank you.
($d=1$ can be taken arbitrarily here. If another value is chosen, the values for $a,b,c$ are scaled by a factor of $d$ and the same plane equations result.)
$$1=\frac{|a+2c+1|}{\sqrt{a^2+b^2+c^2}}\tag1$$ $$a+b-c=-1\tag2$$ $$2a+b+c=-1\tag3$$ $$a+2c=0\tag4$$ Substitute (4) into (1) twice: $$1=\frac{|0+1|}{\sqrt{(-2c)^2+b^2+c^2}}$$ $$1=\frac1{\sqrt{5c^2+b^2}}$$ $$5c^2+b^2=1\tag5$$ Substitute (4) and (5) into (2) to leave an equation in $c$ only: $$-2c+\sqrt{1-5c^2}-c=-1$$ $$\sqrt{1-5c^2}=3c-1$$ $$1-5c^2=9c^2-6c+1$$ $$14c^2-6c=0$$ from which we get $c=0$ or $c=\frac37$. For both cases substitute into (4) and then (2) to get two possible normals: $$a=0,b=-1,c=0$$ $$a=-\frac67,b=\frac27,c=\frac37$$ (2) establishes that $(a,b,c)\cdot P=-1$ in both cases, so we have the plane equations $$-y=-1,\quad-\frac67x+\frac27y+\frac37z=-1$$ which rearrange to the given answers of $$y-1=0,\quad 6x-2y-3z-7=0$$