[tex]\begin{cases}x-6y-3z=4&(1)\\-2x-3z=-8&(2)\\-2x+2y-3z=-14&(3)\end{cases}[/tex]
First eliminate [tex]y[/tex] by adding [tex](1)[/tex] to three times [tex](3)[/tex]. This gives
[tex](x-6y-3z)+3(-2x+2y-3z)=4+3(-14)\iff -5x-12z=-38[/tex]
This reduces the system of one of two equations and two unknowns:
[tex]\begin{cases}-5x-12z=-38&(1)^*\\-2x-3z=-8&(2)^*\end{cases}[/tex]
You can eliminate [tex]z[/tex] by subtracting [tex](1)^*[/tex] and four times [tex](2)^*[/tex] to get
[tex](-5x-12z)-4(-2x-3z)=-38-4(-8)\iff3x=-6\implies x=-2[/tex]
Back-substitute to find [tex]z[/tex] and [tex]y[/tex]. You should end up with [tex](x,y,z)=(-2,-3,4)[/tex].
