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Question

A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A,B,C. Show that the locus of the centroid of triangle ABC is 1x2+1y2+1z2=1p2

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Solution

Let the equation of the plane be
xa+yb+zc=1 ...(i)
where a,b,c are variables.
This plane meets X,Y and Z axes at points A(a,0,0),B(0,b,0) and C(0,0,c) respectively. Let (α,β,γ) be the coordinates of the centroid of ΔABC. Then,
α=a+0+03=a3,β=0+b+03=b3,
γ=0+0+c3=c3 ...(ii)
The plane represented by equation (i) is at a distance 3p from the origin.
3p = Length of perpendicular from (0,0,0) to the plane (i)
3p=0a+0b+0c1(1a)2+(1b)2+(1c)2
3p=11a2+1b2+1c2
19p2=1a2+1b2+1c2 ...(iii)
From (ii), we have
a=3α,b=3β and c=3γ
Substituting the values of a,b,c, in (iii), we obtian
19p2=19α2+1β2+19γ2
1p2=1α2+1β2+1γ2
So, the locus of centroid of triangle is
1p2=1x2+1y2+1z2

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