Three cones and a sphere

Here is a top view of the three cones:

Since the side length of the equilateral triangle is $3$ , by the properties of an equilateral triangle, $AC = \sqrt{3}$ , so $BC = AC - AB = \sqrt{3} - 1$ .

Here is a side view of one of the cones with the ball:

By the Pythagorean Theorem on $\triangle ABD$ , $BD = \sqrt{AD^2 + AB^2} = \sqrt{4^2 + 1^2} = \sqrt{17}$ .

Since $\triangle HEF \sim \triangle ABD$ by AA similarity, $EH = EF \cdot \cfrac{AB}{BD} = \cfrac{3}{2} \cdot \cfrac{1}{\sqrt{17}} = \cfrac{3}{2\sqrt{17}}$ and $FH = EF \cdot \cfrac{AD}{BD} = \cfrac{3}{2} \cdot \cfrac{4}{\sqrt{17}} = \cfrac{6}{\sqrt{17}}$ .

Then $FG = FH - GH = FH - BC = \cfrac{6}{\sqrt{17}} - (\sqrt{3} - 1)$ .

Since $\triangle GFB \sim \triangle ABD$ also by AA similarity, $GB = FG \cdot \cfrac{AD}{AB} = (\cfrac{6}{\sqrt{17}} - (\sqrt{3} - 1)) \cdot \cfrac{4}{1} = \cfrac{24}{\sqrt{17}} - 4\sqrt{3} + 4$ .

The height of the center of the ball is then $EC = EH + HC = EH + GB = \cfrac{3}{2\sqrt{17}} + \cfrac{24}{\sqrt{17}} - 4\sqrt{3} + 4 = \cfrac{3\sqrt{17}}{2} - 4\sqrt{3} + 4 \approx \boxed{3.256}$ .