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Question

A point P moves on a plane xa+yb+zc=1 and O be the origin. A plane through P and perpendicular to OP meets the coordinate axes at A,B and C respectively. Let G be the centroid of the triangle ABC. If the planes through A,B and C parallel to the planes x=0,y=0 and z=0, respectively, intersect at Q, then

A
If OA=OB=OC, then OG=13OP
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B
Locus of Q is
1ax+1by+1cz=1x2+1y2+1z2
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C
Locus of Q is
1a+1b+1c=1x2+1y2+1z2
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D
If OA=OB=OC, then OG=OP
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Solution

The correct option is D If OA=OB=OC, then OG=OP
Let coordinates of P be (h,k,l)
ha+kb+lc=1 (1)OP=h2+k2+l2

Direction cosines of OP is
hh2+k2+l2,kh2+k2+l2,lh2+k2+l2
Equation of the plane passing through P and perpendicular to OP is
hxh2+k2+l2+kyh2+k2+l2+lzh2+k2+l2=h2+k2+l2

hx+ky+lz=h2+k2+l2

A(h2+k2+l2h,0,0)B(0,h2+k2+l2k,0)C(0,0,h2+k2+l2l)OA=OB=OCh=k=l
Centroid, G(h2+k2+l23h,h2+k2+l23k,h2+k2+l23l)G(h,k,l)OG=OP

If coordinates of Q is (e,f,g), then
e=h2+k2+l2hha=h2+k2+l2ae,f=h2+k2+l2kkb=h2+k2+l2bf,g=h2+k2+l2llc=h2+k2+l2cg

Now, 1e2+1f2+1g2=1h2+k2+l2
From (1),
h2+k2+l2ae+h2+k2+l2bf+h2+k2+l2cg=11ae+1bf+1cg=1h2+k2+l21ae+1bf+1cg=1e2+1f2+1g2

Locus of Q is 1ax+1by+1cz=1x2+1y2+1z2

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