    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=16p2Let 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  Suggest Corrections  1      Similar questions
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