Set up the geometry

Let the total height of water column above the ground be $H = 12\text{ m}.$

A hole is made at depth $h$ below the water surface.

Therefore, the hole is at height above the ground: $y = H - h = 12 - h.$

Speed of efflux from the hole

By Torricelli’s theorem, the speed of water emerging from the hole is $v = \sqrt{2gh}.$

The jet comes out horizontally.

Time taken to hit the ground

Since the jet starts horizontally, its initial vertical velocity is zero.

It falls through vertical distance $(12-h)$, so $12-h = \frac{1}{2}gt^2.$

Hence, $t = \sqrt{\frac{2(12-h)}{g}}.$

Horizontal range of the jet

Horizontal distance traveled before striking the ground is $R = vt.$

Substituting $v$ and $t$: $R = \sqrt{2gh}\cdot \sqrt{\frac{2(12-h)}{g}}.$

Simplify: $R = 2\sqrt{h(12-h)}.$

Maximize the range

To maximize $R$, we maximize $h(12-h) = 12h - h^2.$

This is a downward-opening parabola. Its maximum occurs at $\frac{d}{dh}(12h - h^2) = 12 - 2h = 0.$

So, $h = 6\text{ m}.$

Conclusion

The hole should be made at a depth $\boxed{6\text{ m}}$ below the water surface for maximum horizontal range.