Time taken by water to reach on the plane $PQ$ will be

$t=\sqrt{\frac{2\left(\frac{h}{2}+(h-y)\right)}{g}}-\sqrt{\frac{3 h-2 y}{g}}$

Horizontal distance $x$ travelled by the liquid is $x=v t-\sqrt{2 g y} \sqrt{\frac{3 h-2 y}{g}}, \quad x=\sqrt{2 y(3 h-2 y)}$

For $x$ to be maximum, $\frac{d x}{d y}=0$

$\frac{1}{2 \sqrt{2 y(3 h-2 y)}} \times 2(3 h-4 y)=0$

Or           $3 h-4 y-0 ; \quad y=\frac{3 h}{4}$

Hence, $x$ will be maximum at $y=\frac{3 h}{4}=\left(h-\frac{h}{4}\right)$

Where, hole number $2$ is present.