If the areas of triangles formed by a plane with the positive X,Y:Y,Z:Z,X axes respectively are 12, 9, 6 sq. unit respectively then the equation of the plane is

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x4+y6+z3=1

x6+y3+z4=1

x3+y4+z6=1

x3+y6+z4=1

answer is A.

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Detailed Solution

If a,b,c are the intercepts of the required plane then 12ab=12,12bc=9,12ca=6⇒ab=24,bc=18,ca=12a2=abacbc=24×1218=16⇒a=4ab=24,ac=12⇒b=6,c=3 The equation of the plane is x4+y6+z3=1

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