CurveGuy is correct about the method but assumed the initial conditions are all 0. For example \(\displaystyle \dot X = sX-x(0)\). This leads to the linear system:

\(\displaystyle \begin{pmatrix} sX-x(0)\\sY-y(0)\\sZ-z(0) \end{pmatrix} =

\begin{pmatrix} 5 & 5 & 2 \\ -6 & -6 & -5 \\ 6 & 6 & 5 \end{pmatrix} =

\begin{pmatrix} X \\ Y \\ Z \end{pmatrix} \)

where the initial conditions might not be \(\displaystyle 0\). You can solve the system by any favorite method you choose to get the transforms \(\displaystyle X,~Y,~Z\) after which you need to find their inverses.

That \(\displaystyle 6\) with a line above it should be \(\displaystyle -6\). There appears to be a problem with Latex rendering.