equation of plane containing the line of intersection between two planes and a point I am trying to solve the following question:

What is the equation of the plane containing the line of intersection between the planes $x+y+z=1$ and $y+z=0$, as well as the point $P = (2,1,0)$?

The way I tried to do it is to set arbitrary values to find two other points on the plane and then use the normal of a polygon algorithm to calculate the normal to the plane. 
However, this doesn't seem to yield the correct answer ($x-y-z=1$). Can anyone point me in the right direction? 
 A: Let us go for the intersection line first. We have the system of equations
$$
x + y + z = 1 \\
y + z = 0
$$
which can be simplified to
$$
x = 1 \\
y + z = 0
$$
which gives the line
$$
(1, y, -y) = (1,0,0) + (0,1,-1) y
$$
We can extend that line to a plane by
$$
(1,0,0) + (0,1,-1) s + (a, b, c) t
$$
where $s, t \in \mathbb{R}$ and $(a,b,c)$ is a vector we need to choose such that the plane contains $A$:
$$
(2,1,0) 
= (1,0,0) + (0,1,-1) s + (a, b, c) t 
= (1 + at, s + bt, -s + ct)
$$
We are now dealing with two unknown parameters and three unknown components, but have only three equations.
So we try to require $s=0$ and $t=1$ and look if there is a solution:
$$
(2,1,0) 
= (1,0,0) + (a, b, c) \iff
(1,1,0) = (a,b,c)
$$
This gives
$$
(1,0,0) + (0,1,-1)s + (1,1,0) t = (1+t,s+t,-s) \quad (s, t \in \mathbb{R})
$$
as equation for the plane.

Alternate representation of the solution plane:
We can bring this into a single equation in three coordinates, by finding the normal form
$$
n \cdot x = d
$$
where $n$ is a unit normal vector of the plane and $d$ is the (signed) distance to the origin. 
Maybe there is an easier way to do this, but I do not see it right now.
We can calculate a normal vector from the vector product of the plane spanning vectors:
$$
(0,1,-1) \times (1,1,0) = (1, -1, -1)
$$
so a unit normal vector is $n = (1,-1,-1)/\sqrt{3}$.
The distance of the plane to the origin is
$$
d^2 = q = \lVert (1+t, s+t, -s) \rVert^2 = (1+t)^2 + (s+t)^2 + s^2 \\
$$
We look where the gradient vanishes:
$$
0 = \partial q / \partial s
= 2(s+t) + 2s = 4s + 2t
\\
0 = \partial q / \partial t
= 2(1+t) + 2(s+t) = 2s + 4t + 2
$$
This gives the system
$$
4s + 2t = 0 \\
2s + 4t + 2 = 0
$$
or
$$
2s + t = 0 \\
2s + 4t + 2 = 0
$$
or
$$
2s + t = 0 \\
3t + 2 = 0
$$
so
$$
s = 1/3 \\
t = -2/3
$$
so we get
$$
q = (1 - 2/3)^2 + (1/3 - 2/3)^2 + (1/3)^2 
= 1/9 + 1/9 + 1/9 = 3/9 = 1/3
$$
and $d = 1 / \sqrt{3}$.
This gives
$$
\frac{1}{\sqrt{3}} (1,-1,-1) \cdot (x,y,z) = \frac{1}{\sqrt{3}} 
$$
or simply 
$$
(1,-1,-1) \cdot (x,y,z) = 1 \iff \\
x - y - z = 1
$$
A: In addition to the original question, I later did some experimenting and "solved" it using the following approach(please correct me if I am wrong in any area):
$x+y+z=1$
$y+z=0$
Let $z=1$ then solving the two plane equations simultaneously gives $y = -1$ and $x = 1$ so we have a second point which lies on both planes $P_2 = (1,-1,1)$. 
Now let $y=1$ and $x=1$ then simultaneously solving the equation gives $z=-1$ so we now have a third point $P_3 = (1,1,-1)$. 
We can now find a normal to the plane using the normal of a polygon algorithm: 
$(P_3 - P_1)$ x $(P_3-P_1)$ which gives $n = (-2,2,2)$. 
Thus, this can be translated into a plane equation as follows:
$-x + y + z + d = 0$
Substitite any of the 3 points into the above to get $d$. We will use $(2,1,0)$. This gives $d=1$. 
Finally:
$x-y-z=1$
