I came up with about 3.85 inches. I realized that if I constructed an inverted cone tangent to the other cones with its base "level with the [other] cones' tips", then the problem reduces to a more familiar one: determining the radius of a sphere inscribed in a cone. Because the base radius of the cones is given, it follows that the distance between the tips of the cones is 10 inches. Usin' that good ol' Pythagorean Theorem (30-60-90 triangle ratios in this case) and the fact that the base of the constructed cone contains the tips of the three given cones, I calculated the radius of the constructed cone as 10/sqrt(3) inches. Because the axes of the constructed cone and the given cones are all parallel, the constructed cone must be similar to the given cones. I used this similarity relation and the height of the given cones to find the height of the constructed cone: 12/5 = (10/sqrt(3))/h ==> h = 8*sqrt(3). Knowing the base radius and the height of the constructed cone, I carried out the same calculations in the attached image to obtain the exact answer of 20*sqrt(3)/9 inches.