Question

# A variable plane which remains at a constant distance 3p from the origin cuts the co-ordinate axes at A, B, C. The locus of the centroid of triangle ABC is

A

1x+1y+1z=13p

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B

1x2+1y2+1z2=19p2

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C

1x+1y+1z=1p

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D

1x2+1y2+1z2=1p2

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Solution

## The correct option is D 1x2+1y2+1z2=1p2 Let the equation of the plane be xa+yb+zc=1. Then, A=(a,0,0), B=(0,b,0), C=(0,0,c). The centroid of ABC is (a3,b3,c3) Distance of the plane from the origin. 3p=1√1a2+1b2+1c21a2+1b2+1c2=19p2 Let (x,y,z) be the coordinates of the centroid. Then, x=a3⇒a=3x. Also, b=3y, c=3z⇒19x2+19y2+19z2=19p2⇒1x2+1y2+1z2=1p2

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