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Question

If ax+by+cz=d intersects the coordinate axis at A, B, C and the area of the triangle ABC is (a2+b2)2(a2+c2+b2)a2b2(a2+b2). If a2, c2, b2 are in A.P. then find the perpendicular distance of the plane from the origin.

A
23
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B
1
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C
233
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D
2
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Solution

The correct option is C 233
A(da,0,0), B(0,db,0), C(0,0,dc)AB=da^idb^j+0^kBC=0^i+db^jdc^k

Area of the triangle ABC=12|AB×BC|=12d2(a2+b2+c2)abc12d2(a2+b2+c2)abc=(a2+b2)2(a2+c2+b2)a2b2(a2+b2)d22abc=(2c2)(2)abc(2)d2c2=4
dc=2
The distance of plane from origin =0+0+0d(a2+b2+c2)=d3c=233

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