$\begin{array}{l}{\left[\begin{array}{lcc}x & -3 i & 1 \\ y & 1 & i \\ 0 & 2 i & -i\end{array}\right]=6+11 i} \\ & \Rightarrow \mathrm{x}[-\mathrm{i}-(2 \mathrm{i})(\mathrm{i})]-\mathrm{y}[(-3 \mathrm{i})(-\mathrm{i})-(2 \mathrm{i})(1)]+0=6+11 \mathrm{i} \\ & \Rightarrow \mathrm{x}(-\mathrm{i}+2)-\mathrm{y}(-3-2 \mathrm{i})=6+11 \mathrm{i} \\ & \Rightarrow-\mathrm{x} \mathrm{i}+2 \mathrm{x}+3 \mathrm{y}+2 \mathrm{yi}=6+11 \mathrm{i} \\ & \Rightarrow(2 \mathrm{x}+3 \mathrm{y})+(-\mathrm{x}+2 \mathrm{y}) \mathrm{i}=6+11 \mathrm{i} \\ & \therefore 2 \mathrm{x}+3 \mathrm{y}=6 \text { and }-\mathrm{x}+2 \mathrm{y}=11\end{array}$
Solving the equations,
$2 \not x+3 y=6$
$\frac{-2 \not x+4 y=22}{7 y=28} \Rightarrow y=4$
$2 x+3 y=6 \Rightarrow 2 x+12=6$
$\Rightarrow 2 x=-6 \Rightarrow x=-3$
$\therefore \mathrm{x}=-3, \mathrm{y}=4$