Determine a line such that all its points lie at equal distance to three non-parallel planes. I am supposed to determine the parametric equation of a line such that all it's points lie at equal distance to the three planes, 
$$x+2y+2z+3=0$$
$$x-2y+2z-1=0$$
$$2x+y+2z+1=0$$
So far I've been able to determine the point where all the planes intersect, as well as all the intersection lines between the planes individually. However, I am unable to get much further. I've tried determining a line that goes through the intersection point of the planes and another point with equal distance to the planes, but I just end up with a horrible system of equations. I'm thinking that there must be a much simpler way of going about solving this problem.

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} $.
