If $\left|\begin{array}{ccc}6i& -3i& 1\\ 4& 3i& -1\\ 20& 3& i\end{array}\right|=x+iy,$ then :

$\because$ $\left|\begin{array}{ccc}6i& -3i& 1\\ 4& 3i& -1\\ 20& 3& i\end{array}\right|$ 

$=\left|\begin{array}{ccc}6i+4& 0& 0\\ 4& 3i& -1\\ 20& 3& i\end{array}\right|\left(R_1\to R_1+R_2\right)$

$=(6i+4)\left|\begin{array}{cc}3i& -1\\ 3& i\end{array}\right|$

$=\left(6i+4\right)\left(3i^2+3\right)=0$

Put $\left|\begin{array}{ccc}6i& -3i& 1\\ 4& 3i& -1\\ 20& 3& i\end{array}\right|=x+iy$

$\Rightarrow 0+0i=x+iy$

$\Rightarrow x=0,y=0$