Let us assume WLOG that the equations of the planes are under the normalized form :
$$(P_k) \ : \ a_kx+b_ky+c_kz=0 \ \text{with} \ a_k^2+b_k^2+c_k^2=1$$
with independant normal vectors $(a_k,b_k,c_k)$.
(indeed one can change in particular the common point to be the origin of coordinates).
In this case, the distance between a point $M(x,y,z)$ and plane $(P_k)$ is given by :
$$\delta(M,(P_k))=|a_kx+b_ky+c_kz|$$
and the oriented distance by
$$\Delta(M,(P_k))=a_kx+b_ky+c_kz.$$
Therefore, the locus of points that are at equal distance from the 3 planesare defined by the system of equations :
$$\delta(M,(P_1))=\delta(M,(P_2))=\delta(M,(P_3)) \tag{1}$$
(we don't know really know that it is a union of lines even if we have a strong intuition that such is the case).
(1) is equivalent to :
$$\begin{cases}\Delta(M,(P_1)))&=&\color{red}{u}\Delta(M,(P_2))\\
\Delta(M,(P_2)))&=&\color{red}{v}\Delta(M,(P_3))\\
\Delta(M,(P_3)))&=&\color{red}{w}\Delta(M,(P_1))\end{cases}\tag{2}$$
where $\color{red}{(u,v,w)}$ can be $\color{red}{(-,-,-)}$, $\color{red}{(-,-,+)}$, ... to $\color{red}{(+,+,+)}$.
Only some of the combinations will give rise to solutions (in this regard, see the important Edit below).
Let us consider first the example of the 3 planes with equations
$$\begin{cases}2x+2y+z&=&0\\
x-2y-2z&=&0\\
-2x+2y-z&=&0;\end{cases}$$
(Notice that we have taken coefficients such that the common norm is simple : $\sqrt{2^2+2^2+1^2}=3$)
Among the 8 possible sign combinations, 4 of them will give rise to straight lines :
Here is one :
$$\color{red}{(+,+,+)}\begin{cases}\Delta(M,(P_1)))&=&\color{red}{+}\Delta(M,(P_2))\\
\Delta(M,(P_2)))&=&\color{red}{+}\Delta(M,(P_3))\\
\Delta(M,(P_3)))&=&\color{red}{+}\Delta(M,(P_1))\end{cases} \iff \pmatrix{x\\y\\z}=\pmatrix{2r\\-2r\\r}$$
Here are the 3 others :
$$\color{red}{(+,-,-)} \pmatrix{x\\y\\z}=\pmatrix{-3r\\0\\r}$$
$$\color{red}{(-,+,-)} \pmatrix{x\\y\\z}=\pmatrix{-r/3\\0\\r}$$
$$\color{red}{(-,-,+)} \pmatrix{x\\y\\z}=\pmatrix{0\\r\\0}$$
Fig. 1 : The 3 great circles are the intersection of the 3 planes with the unit sphere : they delimitate 8 spherical triangles paired 2 by 2 (each triangle paired with its "antipodal" triangle). In black and red, the 4 "bissecting" lines ; for example the red one intersects the unit sphere in two points which are the equivalent of the incenters of the corresponding spherical triangle.
Here is the SAGE program that I have used :
x,y,z,u,v,w=var('x y z u v w')
e1=2x+2y+z;
e2=x-2y-2z;
e3=-2*x+2y-z;
u=1;v=1;w=1;#to be changed
solve([e1==u*e2,e2==v*e3,e3==w*e1],x,y,z)
(I am much indebted to @Oscar Lanzi who has pointed a big error of mine).
Important Edit :
The fact that there are exactly 4 lines is rather intuitive geometricaly but deserves an algebraic proof.
In fact (2) is equivalent to this matrix equation describing a certain kernel :
$$\underbrace{\pmatrix{1&-u&0\\
0&1&-v\\-w&0&1}}_{M_{u,v,w} \ \text{with rank} \ r}\underbrace{\pmatrix{a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3}}_P \pmatrix{x\\y\\z}=\pmatrix{0\\0\\0};$$
According to the values $\pm1$ of $u,v,w$, the value of rank $r$ will be $2$ or $3$ :
$$\begin{array}{|r|r|r|r|c|r|}
\hline
u&v&w&r&d=3-r\\
\hline
-1&-1&-1&3&0&\text{point}\\
-1&-1&1&2&1&\text{line}\\
-1&1&-1&2&1&\text{line}\\
-1&1&1&3&0&\text{point}\\
1&-1&-1&2&1&\text{line}\\
1&-1&1&3&0&\text{point}\\
1&1&-1&3&0&\text{point}\\
1&1&1&2&1&\text{line}\\
\hline
\end{array}$$