$\begin{aligned} & {\left[\begin{array}{lll}0 & -1 & 4\end{array}\right]\left\{2\left[\begin{array}{cc}4 & 5 \\ 3 & 6 \\ 2 & -1\end{array}\right]+3\left[\begin{array}{cc}4 & 3 \\ 1 & 4 \\ 0 & -1\end{array}\right]\right\}} \\ & =\left[\begin{array}{ll}x & y\end{array}\right] .\end{aligned}$
$\begin{aligned} & \therefore \quad\left[\begin{array}{lll}0 & -1 & 4\end{array}\right]\left\{\left[\begin{array}{ll}8 & 10 \\ 6 & 12 \\ 4 & -2\end{array}\right]+\left[\begin{array}{cc}12 & 9 \\ 3 & 12 \\ 0 & -3\end{array}\right]\right\}=\left[\begin{array}{ll}x & y\end{array}\right] \\ & \therefore \quad\left[\begin{array}{lll}0 & -1 & 4\end{array}\right]\left[\begin{array}{ll}20 & 19 \\ 9 & 24 \\ 4 & -5\end{array}\right]=\left[\begin{array}{ll}x & y\end{array}\right] \\ & \therefore \quad\left[\begin{array}{ll}0-9+16 & 0-24-20\end{array}\right]=\left[\begin{array}{ll}x & y\end{array}\right]\end{aligned}$
$\therefore \quad\left[\begin{array}{ll}7 & -44\end{array}\right]=[x, y]$
$\therefore \quad$ By equality of matrices, we get
$x=7, y=-44$
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$\left[\begin{array}{lll}9 & \sqrt{2} & -3\end{array}\right]$
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by using properties.