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.
 A: 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} $.
A: First, how to build the enveloping cone of a sphere.
Given a sphere $S\to \lVert p-p_c\rVert=r$ and a line $L\to p = p_0+\mu \vec v$ with $p = (x,y,z)^{\dagger}$ and $p_0$ the vertex, their intersection $S\cap L$ is obtained by solving for $\mu$
$$
\lVert p_0-p_c +\mu \vec v\rVert^2=r^2
$$
so
$$
\mu = \frac{-2(p_0-p_c)\cdot\vec v\pm\sqrt{4\left((p_0-p_c)\cdot\vec v\right)^2-4\left(\lVert p_0-p_c\rVert^2-r^2\right)\lVert\vec v\rVert^2}}{2\lVert\vec v\rVert^2}
$$
but if tangency is required then
$$
\left((p_0-p_c)\cdot\vec v\right)^2-\left(\lVert p_0-p_c\rVert^2-r^2\right)\lVert\vec v\rVert^2 = 0
$$
but $\vec v = \frac{1}{\mu}(p-p_0)$ so after substitution we have the cone equation
$$
\left((p_0-p_c)\cdot(p-p_0)\right)^2-\left(\lVert p_0-p_c\rVert^2-r^2 \right)\lVert p-p_0\rVert^2=0
$$
Given three planes
$$
\begin{array}{rcl}
\Pi_1\to (p-p_1)\cdot \vec n_1 = 0\\
\Pi_2\to (p-p_2)\cdot \vec n_2 = 0\\
\Pi_3\to (p-p_3)\cdot \vec n_3 = 0
\end{array}
$$
with 
$$
\begin{array}{rcl}
p & =& (x,y,z)^{\dagger}\\
p_1 & = & (-3,0,0)^{\dagger}\\
p_2 & = & (1,0,0)^{\dagger}\\
p_3 & = & (0,-1,0)^{\dagger}\\
\vec n_1 & = &(1,2,2)^{\dagger}\\
\vec n_2 & = & (1,-2,2)^{\dagger}\\
\vec n_3 & = & (2,2,2)^{\dagger}\\
p_0 & = & (1,-1,-1)^{\dagger}
\end{array}
$$
with $\Pi_1\cap\Pi_2\cap\Pi_3= p_0$ a generic right cone tangent to the three planes and with vertex at $p_0$ can be written as
$$
C\to \left((p_0-p_c)\cdot(p-p_0)\right)^2-\left(\lVert p_0-p_c\rVert^2-r^2 \right)\lVert p-p_0\rVert^2 = 0
$$
Now considering $C$ and $\Pi_1$ if they are tangent with $p_1^*$ as tangency point, we have
$$
\nabla C(p_1^*) = \lambda_1\vec n_1,\ \ 
(p_1^*-p_1)\cdot\vec n_1 = 0
$$
or
$$
\begin{array}{rcl}
((p_0-p_c)\cdot(p_k^*-p_0))(p_0-p_c)-\left(\lVert p_0-p_c\rVert^2-r^2 \right)(p_k^*-p_0)& =  &\lambda_k\vec n_k\\
(p_k^*-p_k)\cdot \vec n_k & = & 0
\end{array}
$$
giving four equations in the unknowns $x_k^*,y_k^*,z_k^*,\lambda_k$
Here the tangency points $p_k^*$ are located along an infinite line so
for each $k$ we have
$$
M_k(p_c,r)\cdot (x_k^*,y_k^*,z_k^*,\lambda_k)^{\dagger} = b_k
$$
should be such that $\det(M_k) = f_k(p_c,r) = 0$
and the relationship given by $f_k(p_c,r) = 0$ furnish the equidistant lines in the form
$$
L_j\to p_c = p_0+\omega_j r
$$
For the actual example we have
$$
\begin{array}{rcl}
f_1(p_c.r) & = &  (3r-3-x_c-2y_c-2z_c)(3r+3+x_c+2y_c+2z_c)\\
f_2(p_c,r) & = & (3 r+1 - x_c + 2 y_c - 2 z_c)(3 r-1 + x_c - 2 y_c + 2 z_c) \\
f_3(p_c,r) & = & (3 r-1 - 2 x_c - y_c - 2 z_c) (3 r+1 + 2 x_c + y_c + 2 z_c)
\end{array}
$$
arranging and solving for $p_c$ we get one
$$
\begin{array}{rcl}
3 r - x_c - 2 y_c - 2 z_c & = & 3\\
3 r - x_c + 2 y_c - 2 z_c & = & -1\\
3 r - 2 x_c - y_c - 2 z_c & = & 1
\end{array}
$$
$$
p_c = p_0 + \frac{3}{2}(0,0,1)^{\dagger}r
$$
and another one by solving
$$
\begin{array}{rcl}
3 r + x_c + 2 y_c + 2 z_c & = & -3\\
3 r + x_c - 2 y_c + 2 z_c & = &  1\\
3 r + 2 x_c + y_c + 2 z_c & = & -1
\end{array}
$$
$$
p_c = p_0 + \frac{3}{2}(1,1,-\frac{1}{2})^{\dagger}r
$$
and another
$$
\begin{array}{rcl}
3 r - x_c - 2 y_c - 2 z_c & = & 3\\
3 r - x_c + 2 y_c - 2 z_c & = & -1\\
3 r + 2 x_c + y_c + 2 z_c & = & -1
\end{array}
$$
giving
$$
p_c = p_0 + (-6,0,\frac 92)^{\dagger}r
$$
etc.
Attached a plot showing the three planes and the three cones associated to the three center lines. Obviously, those center lines are the sought lines,

