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
$$
(1,-1,-1) \cdot (x,y,z) = 1 \iff \\
x - y - z = 1
$$
You can instead minimize the squared distance to simplify taking derivatives since squaring is an increasing transformation (minimizing the distance is the same as minimizing the distance squared).
$$\mathcal{L} = x^2 +y^2 + z^2 + \lambda (2z+y-12) + \mu (x+y-6)$$
Then find $x,y,z$ minimizing this in terms of the lagrange multipliers. Then substitute those values back into the lagrangian above, and finally optimize the lagrange multipliers.
Best Answer
You can make the equations more tractable by using the square of the distance as the objective function instead. However, there’s also a fairly straightforward solution method based on geometric considerations.
The minimal distance to the line is measured along a direction perpendicular to the line, so the nearest point on the line to the origin is the intersection of the line with its perpendicular plane through the origin. You have $(2,-2,-1)^T$ as a direction vector for the line, so an equation of this plane is $2x-2y-z=0$. You will end up with a simple linear equation in $t$ to solve.