1. **Problem Statement:** Given three chords AB, CD, and EF of equal length in a circle with center O, and points L, M, N on AB, CD, EF respectively such that OL \perp AB, OM \perp CD, and ON \perp EF. It is given that \angle QPR = 79^\circ. Find \angle QOR. 2. **Key Concepts:** - The perpendicular from the center of a circle to a chord bisects the chord. - The angle subtended by chords at the center relates to the angles subtended at other points. - The chords AB, CD, EF are equal, so their perpendicular distances from the center O are equal. 3. **Step-by-step Solution:** - Since AB = CD = EF, and OL, OM, ON are perpendiculars from O to these chords, OL = OM = ON. - Points L, M, N lie on the chords such that OL \perp AB, OM \perp CD, ON \perp EF. - The points P, Q, R lie on the circle such that \angle QPR = 79^\circ. - The angle \angle QOR at the center subtended by the arcs corresponding to chords QP and PR is twice the angle \angle QPR at the circumference (by the circle theorem: angle at center is twice angle at circumference subtending the same arc). - Therefore, \angle QOR = 2 \times 79^\circ = 158^\circ. 4. **Final Answer:** $$\boxed{\angle QOR = 158^\circ}$$