Which of the three-dimensional primitive lattices has a unit cell where none of the internal angles is \(90^{\circ}\) ? (a) Orthorhombic, (b) hexagonal, (c) rhombohedral, (d) triclinic, (e) both rhombohedral and triclinic.

Here's a brief overview of the given lattice types and their unit cell properties: (a) Orthorhombic: Lot has three cell edges (a, b, and c) with different lengths and all internal angles equal to \(90^{\circ}\). (b) Hexagonal: Lattice has three cell edges of equal length (a=a=b), and only two internal angles equal to \(90^{\circ}\) while the other is equal to \(120^{\circ}\). (c) Rhombohedral: Lattice has three cell edges of equal length (a=a=b) and all internal angles are equal but not equal to \(90^{\circ}\). (d) Triclinic: Lattice has three cell edges (a, b, and c) with different lengths, and no internal angles are equal to \(90^{\circ}\). Now, let's analyze which of these options meet the requirement of having no internal angle equal to \(90^{\circ}\).

Looking at the properties mentioned earlier, we can rule out option (a) - Orthorhombic and option (b) - Hexagonal because they have at least one internal angle equal to \(90^{\circ}\). Option (c) - Rhombohedral lattice has all internal angles equal but not equal to \(90^{\circ}\), which satisfies the requirements of this problem. Option (d) - Triclinic lattice has no internal angles equal to \(90^{\circ}\), which also meets the requirements of this problem. Therefore, both rhombohedral and triclinic lattices fulfill the given condition.

Since both option (c) - Rhombohedral and option (d) - Triclinic lattices fulfill the given condition, the correct answer is (e) both rhombohedral and triclinic.