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

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 $\frac{x}{1}=\frac{y}{1}=\frac{z}{1}$. 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 $1(x-a)+1(y-a)+1(z-a)=0$ $(\because$ OP is normal to the plane $)$ i.e., $x+y+z=3 a$ $\Rightarrow \frac{x}{3 a}+\frac{y}{3 a}+\frac{z}{3 a}=1$ Intercepts on axes are $3 \mathrm{a}, 3 \mathrm{a}$ and $3 \mathrm{a}$, therefore sum of reciprocals of these intercepts. $=\frac{1}{3 a}+\frac{1}{3 a}+\frac{1}{3 a}=\frac{1}{a}$