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 D1a 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.