Answer
$\{(z+6, z+2, z),\ z\in \mathbb{R}\}$
Work Step by Step
The equations for the intersections:
$\left\{\begin{array}{ll}
I_{1}: & x-y=4\\
I_{2}: & x-z=6\\
I_{3}: & y-z=2
\end{array}\right.$
Augmented matrix:
$\left[\begin{array}{lllll}
1 & -1 & 0 & | & 4\\
1 & 0 & -1 & | & 6\\
0 & 1 & -1 & | & 2
\end{array}\right]\left(\begin{array}{l}
.\\
\leftarrow(R_{2}-R_{1}).\\
.
\end{array}\right)$
$\left[\begin{array}{lllll}
1 & -1 & 0 & | & 4\\
0 & 1 & -1 & | & 2\\
0 & 1 & -1 & | & 2
\end{array}\right]\left(\begin{array}{l}
.\\
.\\
\leftarrow(R_{3}-R_{2}).
\end{array}\right)$
$\left[\begin{array}{lllll}
1 & -1 & 0 & | & 4\\
0 & 1 & -1 & | & 2\\
0 & 0 & 0 & | & 0
\end{array}\right]\left(\begin{array}{l}
\leftarrow(R_{1}+R_{2}).\\
.\\
.
\end{array}\right)$
$\left[\begin{array}{lllll}
1 & 0 & -1 & | & 6\\
0 & 1 & -1 & | & 2\\
0 & 0 & 0 & | & 0
\end{array}\right]$
The system is consistent,
with infinitely many solutions.
Taking z as parameter,
express x and y in terms of z.
$x-z=6 \Rightarrow x=z+6$
$y-z=2 \Rightarrow y=z+2$
Solution set: $\{(z+6, z+2, z),\ z\in \mathbb{R}\}$.