3.26. A plane is drawn through the height of an equilateral triangle with side a, perpendicular to the plane of the triangle, and in this plane a line $l$ is taken, parallel to the height of the triangle. Find the volume of the body obtained by rotating the triangle around the line $l$.
3.26. First, note that when a segment of length $2 d$ is rotated (in a plane) about a point lying on the perpendicular bisector of this segment and at a distance $x$ from the segment, a ring is formed with an inner radius of $x$ and an outer radius of $\sqrt{x^{2}+d^{2}}$; the area of this ring is $\pi d^{2}$, i.e., it does not depend on $x$. Therefore, the intersection of the given body with a plane perpendicular to the axis of rotation represents a ring whose area does not depend on the position of the line $l$. Hence, it is sufficient to consider the case where the axis of rotation is the height of the triangle. In this case, the volume of the body of rotation - a cone - is $\pi a^{3} \sqrt{3 / 24}$.