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Question

P is a fixed point (a,a,a) on a line through the origin equally inclined to the axes, then any plane through P perpendicular to OP, makes intercepts on the axes, the sum of whose reciprocals is equal to

A
a
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B
32a
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C
3a2
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D
1a
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Solution

The correct option is D 1a
Since the line is equally inclined to the axes and passes through the origin, its direction ratios are 1, 1, 1. So its equation is x1=y1=z1
A point P on it is given by (a, a, a). So equation of the plane through P(a,a,a) and perpendicular to OP is 1(x-a)+1(y-a)+1(z-a)=0
[ OPis normal to the plane]
i.e. x+y+z =3a or x3a+y3a+z3a=1
Intercepts on axes are 3a, 3a, 3a. Therefore sum of reciprocals of these intercepts
=13a+13a+13a=33a=1a.

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