I think it is not possible!
In order to find if there is an intersection point we have to find the equations that describe the trajectories of its stream of water and demand that they have a common point.
Thus let one hole to live at height h from the ground while the surface of the water in the cylinder lives at height H (at some instance of time). Furthermore we assume that the velocity of the surface is negligible with respect to the velocity of the water leaving from the hole. Then from
Bernoulli's law we have
\rho\,g\,H=\frac{1}{2}\,\rho\,v^2+\rho\,g\,h\Rightarrow v^2=2\,g\,(H-h)
When the water leaves the hole it feels only its weight, so the equations of motion are
\left\{\begin{array}{l}x=v\cdot t\\ y=h-\frac{1}{2}\,g\,t^2\end{array}\Rightarrow y=h-\frac{g}{2\,v^2}\,x^2\Rightarrow y=h-\frac{x^2}{4\,(H-h)} \quad (1)
I choosed my coordinate system having its origin at the bottom corner of the cylinder, right below the holes.
Now take the first two holes living at heights h_1,\,h_2 respectively. In order the trajectories to have a common point they must both fulfill equation (1) for some coordinates (x_o>0,y_o>0), i.e.
\left{\begin{array}{l} y_o=h_1-\frac{x_o^2}{4\,(H-h_1)}\\y_o=h_2-\frac{x_o^2}{4\,(H-h_2)}\end{array}\Rightarrow \left\{\begin{array}{l}x_o=2\,\sqrt{(H-h_1)(H-h_2)} \\ y_o=h_1+h_2-H \end{array}
Now for the 3rd hole in height h_3 equation (1) merely defines h_3 if we plug in it the above values of (x_o,y_o). Thus we arrive to
y_o=h_3-\frac{x_o^2}{4\,(H-h_3)}\Rightarrow \frac{(h_3-h_1)(h_3-h_2)}{H-h_3}=0\Rightarrow h_3=h_1 \quad \text{or} \quad h_3=h_2
Thus the 3rd hole must be identified with the 1st or the 2nd hole, which means that we can't have three holes with a common intersetion point for their streams.
