Point D is equidistant from all vertices of a regular triangle and is located at a distance of 3 cm from its plane

Point D is equidistant from all vertices of a regular triangle and is located at a distance of 3 cm from its plane. The height of the triangle is 6 cm. Find the distance from point D to the vertices of the triangle.

Since D is equidistant from the vertices of the triangle, its projection onto the plane is the center of the circumscribed circle of the triangle.
The center of the circumcircle and the center of the inscribed circle at a regular triangle coincide and lie at the intersection of the heights of the triangle.
The radius of the circumscribed circle is 2 times greater than that of the inscribed circle.
The sum of the radius of the inscribed and circumscribed circles is equal to the height of the triangle.
Therefore, the radius of the circumscribed circle is 4 cm (the radius of the inscribed circle is 2 cm).
The distance from D then the vertices are calculated by the Pythagorean theorem.
R = √ (4 * 4 + 3 * 3)
Answer: 5 cm.