A sphere of diameter 12 cm is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 359 cm. Find the diameter of the cylindrical vessel.

Increase in the height of the cylinder due to the sphere, h = 359 cm. (Assuming this is a typo and should be a smaller value, let's proceed with h = 3.59 cm for a realistic solution.)
Diameter of sphere = 12 cm
Radius of the sphere, R = 6 cm
Rise in the volume of water in cylinder = Volume of sphere
πr²h = (4/3)πR³
where r is the radius of the cylindrical vessel and h is the rise in water level.
Substituting the values:
πr²(3.59) = (4/3)π(6)³
3.59r² = (4/3)(216)
3.59r² = 288
r² = 288 / 3.59
r² ≈ 80.22
r ≈ √80.22
r ≈ 8.96 cm
Therefore, the diameter of the cylindrical vessel = 2r ≈ 2 * 8.96 cm ≈ 17.92 cm.

If the height increase was actually 359cm (which seems unlikely given the sphere's size), the calculation would be:
πr²(359) = (4/3)π(6)³
359r² = (4/3)(216)
359r² = 288
r² = 288 / 359
r² ≈ 0.802
r ≈ √0.802
r ≈ 0.896 cm
Therefore, the diameter of the cylindrical vessel would be approximately 1.79 cm. This result seems unrealistic given the problem description. It's highly likely there's a typo in the given 'h' value (359cm). We should use a more realistic value for h, such as 3.59 cm as shown in the first calculation for a meaningful answer.