To get a normal vector to the paraboloid at a point (x,y,z), we can take the gradient $\nabla f(x,y,z)=-2xi-2yj+k$. Since we want the tangent plane at the point to be parallel to the plane $x+y+z=1$, the normal vector $\nabla f(x,y,z)=-2xi-2yj+k$ has to be parallel to the vector $i+j+k$ (since this is a normal vector to $x+y+z=1$). This means that $-2xi-2yj+k$ must be a constant multiple of $i+j+k$, so $-2xi-2yj+k=c(i+j+k)$ for some constant c. Then
$-2x=c$, $-2y=c$, and $1=c$, so $x=-1/2$ and $y=-1/2$. Therefore $z=x^2+y^2=1/4+1/4=1/2$ at the point of tangency, and the tangent plane has equation $x+y+z=-1/2$
at this point.
The coefficients of the variables give you the coordinates of a pair $v_1,v_2$ of vectors that are perpendicular to each plane. In you example the vectors $(1,2,-2)$ and $(0,1,0)$.
If the vectors were proportional it would mean that the planes are parallel and therefore the set of points equidistant from the two planes is the parallel plane in between. The quation of this plane would be $v_1\cdot (x,y,z)=(a+b)/2$ where $v_1\cdot(x,y,z)=a$ and $v_2\cdot(x,y,z)=b$ are the equations of the two given planes.
In your case the two vectors are not proportional. Therefore the planes intersect. In this case the set of points equidistant to the two planes are two planes that bisect the angle in between the two given planes.
To get these planes you just need the normal vectors to them. The displacement can then be found be imposing that they pass through some point of the intersection of the two given planes.
The normal vectors that we need are not the cross product of $v_1,v_2$. Such a cross product would point parallel to the line of intersection of the two planes.
What we need is the vectors that bisect the angles between $v_1$ and $v_2$. This bisector is easier to get if we first normalize $v_1$ and $v_2$, because then we just need to bisect a rhombus and the diagonals do this. We get $\frac{v_1}{\|v_1\|}+\frac{v_2}{\|v_2\|}$ and $\frac{v_1}{\|v_1\|}-\frac{v_2}{\|v_2\|}$ as the vectors we are looking for.
Best Answer
Set two of the variables to $0$ and solve for the remaining one. For example,
$$x=y=0\implies 6x+y+9z=9z=54\implies z=6$$
so the $z$-intercept is the point $(0,0,6)$.