Put the two given equations together in a system, set $x=0$ (say) and solve for $y,z$ to get an actual point on the intersection line. Accompanied by its direction vector and the outlying point $P$, we can determine three points from the desired plane and subsequently determine an equation for it.
Dostre's edit (see comments)
$x=0:\begin{cases} y-2z-3=0 \\ -y+z-2=0 \end{cases}$ ;$\;\;\;\;$ add them up and you get:
$\;\;\;\;\;\;\;\;\;\;\begin{cases} -z-5,\;\;z=-5\\ -y+z-2=0,\;y=-7 \end{cases}\Rightarrow$Point $ (0,-7,-5)$, call it Q, is on the line of intersection.
So now we have two points $P(-1,4,2)$ and $Q(0,-7,-5)$ on our desired plane and vector w thats is parallel to the desired plane. In order to find the equation of the desired plane we need a vector that is normal to it.We can find that normal vector by taking cross product of two vectors that are parallel to the desired plane. We already have w so the other vector will be
*PQ*$<0-(-1),-7-4,-5-2>=<1,-11,-7>$
Now normal vector to desired plane will be the cross product of w and PQ:
PQ x w=$\begin{vmatrix} i & j & k \\ 1 & -11 & -7 \\ 1 & 10 & 6 \end{vmatrix}=i\begin{vmatrix} -11 & -7 \\ 10 & 6 \end{vmatrix}-j\begin{vmatrix} 1 & -7 \\ 1 & 6 \end{vmatrix}+k\begin{vmatrix} 1 & -11\\ 1 & 10 \end{vmatrix}=4i-13j+21k$
$4i-13j+21k=<4,-13,21>$ is the vector normal to the desired plane.
Once you know that the plane contains the vector $r$ and the vector $n$, you can determine its normal vector $k$. It also must contain the intersection line, hence contain at least one point of that line. So you need to find a single point $P$ of that line. I suggest you do so by setting $x = 0$ and solving for $y$ and $z$. Once you've done that, the plane equation is
$$
(X - P) \cdot k = 0.
$$
But as I noted in my comment, you've computed $r$ incorrectly, so you need to fix that.
Best Answer
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 $$ (1,-1,-1) \cdot (x,y,z) = 1 \iff \\ x - y - z = 1 $$