The correct option is C 1x2+1y2+1z2=16p2
Let points are : A(a,0,0),B(0,b,0) and C=(0,0,c)
So, equation of plane using intercept form :
xa+yb+zc=1⋯(i)
Converting it into normal form :
⇒xa√1a2+1b2+1c2+yb√1a2+1b2+1c2+zc√1a2+1b2+1c2=1√1a2+1b2+1c2
So, perpendicular distance p=1√1a2+1b2+1c2⋯(ii)
Now, centroid of tetrahedron OABC:
(x,y,z)≡(a4,b4,c4)⇒a=4x,b=4y,c=4z
putting the above values in (ii), we get
116x2+116y2+116z2=1p2⇒1x2+1y2+1z2=16p2