You did right the first step: if the planes have a common point the line shall pass through it.
However the planes do not need in general to have a common point.
The concept to apply is that, given two planes, the points equi-distant from them lie on one of the two planes bisecting the angles between the given planes. If these are parallel, then there is only one bi-secting plane (that in between) if they are distinct, or infinite if they are coincident .
Then, given two non-parallel planes $\pi_1=0, \quad \pi_2=0$ with unitary normal vectors $\bf n_1,\; \bf n_2$, the bisecting plane :
- belongs to the sheaf $\lambda \pi_1+ \mu \pi_2=0$
- has a unitary normal vector proportional to $\bf n_1+\bf n_2$ (external angle) or $\bf n_1-\bf n_2$ (internal angle).
Thus, having three planes,
- take two couples of them (e.g. $\pi_1,\,\pi_2$ and $\pi_2,\,\pi_3$)
- determine the four (or less, if you do not use homogeneous coordinates) bisecting planes $\pi_{1,2,a},\, \pi_{1,2,b}, \, \pi_{2,3,a},\, \pi_{2,3,b}$
- any line given by the crossing of two planes $\pi_{1,2,x}$ & $\pi_{2,3,y}$ will have $d_1=d_2\,\& \,d_2=d_3$.
In conclusion, for three non-parallel planes we have $4$ equi-distant lines. Less than that if some of the planes are parallel.
The above when the distance is measured in absolute terms. If on each plane
a direction of its normal is chosen as to measure the distance in algebraic ($\pm$) terms,
then the line is unique (or does not exist).
To better visualize the whole situation, let's reduce the problem in 2D.
Given three non-parallel lines, thus a non-degenerate triangle made by them,
the points that have the same absolute distance from the three lines are $4$:
the $C_k$ shown in the sketch.
Coming to your particular case, the three planes are concurrent
in one point: the system has only one solution $P=(1,-1,-1)$.
The unit normals to the planes are
$$
\eqalign{
& {\bf n}_{\,1} = 1/3\left( {1,2,2} \right) \cr
& {\bf n}_{\,2} = 1/3\left( {1, - 2,2} \right) \cr
& {\bf n}_{\,3} = 1/3\left( {2,1,2} \right) \cr}
$$
Four bisecting planes are
$$
\eqalign{
& \pi _{1,2,a} = x + 0y + 2z + 1 = x + 2z + 1 = 0 \cr
& \pi _{1,2,b} = 0x + 2y + 0z + 2 = y + 1 = 0 \cr
& \pi _{2,3,a} = {3 \over 2}x - {1 \over 2}y + 2z + 0 = 3x - y + 4z = 0 \cr
& \pi _{2,3,b} = - {1 \over 2}x - {3 \over 2}y + 0z - 1 = x + 3y + 2 = 0 \cr}
$$
they are of course all passing through the point P.
Then starting and taking $\pi _{1,2,a} $ and $ \pi _{2,3,a} $, the cross product of their normals is $(2,2,-1)$.
Therefore a first line is
$$
l_{\,1} :\;{{x - 1} \over { 2}} = {{y + 1} \over 2} = {{z + 1} \over { - 1}} = t
$$
In fact, inserting its generic point $P_1(t)=(1+2t,-1+2t,-1-t)$ into the (normalized) equations of the three planes
we get $4/3t(1,-1,1)$.
And you can check that you get analogue results
with the other three lines obtained by the combination of
$\pi _{1,2,x} $ and $ \pi _{2,3,y} $.
Best Answer
let $M=(3t,1-t,2-2t) $ be a point of the line.
its distance to $z-$axis is
$$\sqrt {9t^2+(1-t)^2} =$$ $$\sqrt {10t^2-2t+1} $$
the derivative inside is $$20t-2$$
the minimum is attained for $t=\frac {1}{10} $ which gives the distance $\frac {3}{\sqrt {10}} $.