For an equilateral triangle, the orthocentre coincides with the centroid. Let $D$ be the foot of perpendicular from $O$ (origin) to line $BC:x+2\sqrt{2}y-4=0$.

$OD=\frac{|0+0-4|}{\sqrt{1+8}}=\frac{4}{3}$.

In an equilateral triangle, the centroid divides the median in the ratio $AO:OD=2:1$, so $AO=\frac{8}{3}$ and the full median $AD=\frac{8}{3}+\frac{4}{3}=4$.

The line $AD$ is perpendicular to $BC$ and passes through $O$. Slope of $BC$ is $-\frac{1}{2\sqrt{2}}$, so slope of $AD$ is $2\sqrt{2}$. This forces $\beta =2\sqrt{2}\alpha$. ...(1)

Distance from $A(\alpha ,\beta )$ to line $BC$ equals $AD=4$:

$\frac{|\alpha +2\sqrt{2}\beta -4|}{3}=4\Rightarrow |\alpha +2\sqrt{2}\beta -4|=12$.

Using $\beta =2\sqrt{2}\alpha$: $|\alpha +8\alpha -4|=12\Rightarrow |9\alpha -4|=12\Rightarrow \alpha =\frac{16}{9}$ or $\alpha =-\frac{8}{9}$.

$A$ and $O$ must lie on the same side of line $BC$ (the orthocentre side). Testing both: $\alpha =\frac{16}{9}$ gives a sign opposite to $O$; reject. So $\alpha =-\frac{8}{9}$, $\beta =-\frac{16\sqrt{2}}{9}$.

$|\alpha +\sqrt{2}\beta |=\left|-\frac{8}{9}-\frac{32}{9}\right|=\frac{40}{9}\approx 4.44$.

$\lfloor 4.44\rfloor =4$. Option (4).