VITEEE 2013 Paper

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

x

1

=

y

1

=

z

1

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
(Since OP is normal to the plane)
i.e.,x + y+z = 3a
⇒

x

3a

+

y

3a

+

z

3a

= 1
Intercepts on axes are 3 a, 3 a and 3 a, therefore sum of reciprocals of these intercepts.
=

1

3a

+

1

3a

+

1

3a

=

1

a