$\begin{bmatrix} 1 &2 &3 \\ 1& 3 &4 \\ 2& 2 &3 \end{bmatrix} \begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 6\\8 \\ 12 \end{bmatrix}$
$x+2y+3z=6........(1)$
$x+3y+4z=6........(2)$
$2x+2y+3z=12........(3)$
Apply (3)-(1)
x=6
Put value of x in any of 2 equations.Let's take (1) and (2)
$2y+3z=0$
$3y +4z=2$
$y=6 , z=-4$
Hence,Option(C) $x=6,y=6,z=-4$.