The ordered triad $(x,y,z)$ which satisfies the system of linear equations is:
$x+y+z=6$
$x-y+z=2$
$3x+2y-4z=-5$

The correct option is A $(1,2,3)$
$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ 1& -1& 1\\ 3& 2& -4\end{array}\right|\text{Applying}{C}_2\to C_2-C_1,C_3\to C_3-C_1=\left|\begin{array}{ccc}1& 0& 0\\ 1& -2& 0\\ 3& -1& -7\end{array}\right|=14{\Delta }_x=\left|\begin{array}{ccc}6& 1& 1\\ 2& -1& 1\\ -5& 2& -4\end{array}\right|\text{Applying}{R}_2\to R_2-R_1,R_3\to R_3+4R_1=\left|\begin{array}{ccc}6& 1& 1\\ -4& -2& 0\\ 19& 6& 0\end{array}\right|=14{\Delta }_y=\left|\begin{array}{ccc}1& 6& 1\\ 1& 2& 1\\ 3& -5& -4\end{array}\right|=\left|\begin{array}{ccc}1& 6& 1\\ 0& -4& 0\\ 0& -23& -7\end{array}\right|=28$
and ${\Delta }_z=\left|\begin{array}{ccc}1& 1& 6\\ 1& -1& 2\\ 3& 2& -5\end{array}\right|=\left|\begin{array}{ccc}1& 1& 6\\ 0& -2& -4\\ 0& -1& -23\end{array}\right|=42$
Hence by Cramer's rule,
$x=\frac{{\Delta }_x}{\Delta }=1,y=\frac{{\Delta }_y}{\Delta }=2,z=\frac{{\Delta }_z}{\Delta }=3$