*Topic: Sphere and Miscellaneous Problems on Sphere*
If (2,3,5) is one end of a diameter of the sphere x2+y2+z2−6x−12y−2z+20=0, then the coordinates of the other end of the diameter are
Solutions · 1
Official solution
Centre =(3,6,1) (from −u,−v,−w with 2u=−6,2v=−12,2w=−2).
\[\frac{\alpha+2}2=3,\ \frac{\beta+3}2=6,\ \frac{\gamma+5}2=1\implies (\alpha,\beta,\gamma)=(4,9,-3).\]