Let the two fixed points be A(−3,1,2) and B(1,−2,4). Let P(x,y,z) be the point at which the line segment AB subtends a right angle. Then, ∠APB=90∘ ⇒‌‌PA⋅PB=0 Now, PA=(−3−x,1−y,2−z) PB=(1−x,−2−y,4−z) So, PA⋅PB=(−3−x)(1−x)+(1−y)(−2−y)+(2−z)(4−z)=0 ⇒−3+3x−x+x2−2−y+2y+y2+8−2z−4z+z2=0 ⇒x2+2x+y2+y+z2−6z+3=0 ⇒x2+y2+z2+2x+y−6z+3=0
