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Question

A variable plane intersects the coordinate axes at A,B,C and is at a constant distance p from O(0,0,0). Then the locus of the centroid of the tetrahedron OABC is

A
1x2+1y2+1z2=1p2
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B
1x2+1y2+1z2=4p2
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C
1x2+1y2+1z2=16p2
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D
1x2+1y2+1z2=16p2
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Solution

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 :
xa1a2+1b2+1c2+yb1a2+1b2+1c2+zc1a2+1b2+1c2=11a2+1b2+1c2
So, perpendicular distance p=11a2+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=1p21x2+1y2+1z2=16p2

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