A plane passes through a fixed point (a, b, c) and cuts the axes in A, B, C. The locus of a point equidistant from origin, A, B and C must be

Let A, B, C be  (α,0,0), (0, β, 0) and (0,0,γ) then the plane ABC is xα+yβ+zγ=1Since it always passes through a, b, c aα+bβ+cγ=1If p is (u,v,w) then OP2=AP2=BP2=CP2⇒u2+v2+w2=(u−α)2+v2+w2=.....⇒ α=u2, β=v2, γ=w2On putting α, β, γ in (1) we get au+bv+cw=2⇒locus of (u, v, w) is ax+by+cz=2