A point P(a, a, a) lies on a line passing thr | vedclass

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(A) The line passes through the origin and makes equal angles with the axes,so its direction ratios are $(1, 1, 1)$.The equation of the line is $\frac

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(A) The line passes through the origin and makes equal angles with the axes,so its direction ratios are $(1, 1, 1)$.The equation of the line is $\frac{x}{1} = \frac{y}{1} = \frac{z}{1}$.The point $P$ is $(a, a, a)$.The plane passes through $P(a, a, a)$ and is perpendicular to $OP$. The direction ratios of the normal to the plane are the same as the direction ratios of the line $OP$,which are $(1, 1, 1)$.The equation of the plane is $1(x - a) + 1(y - a) + 1(z - a) = 0$.Simplifying this,we get $x + y + z = 3a$.Dividing by $3a$,we get $\frac{x}{3a} + \frac{y}{3a} + \frac{z}{3a} = 1$.The intercepts on the axes are $3a, 3a, 3a$.The sum of the reciprocals of the intercepts is $\frac{1}{3a} + \frac{1}{3a} + \frac{1}{3a} = \frac{3}{3a} = \frac{1}{a}$.

$A$ point $P(a, a, a)$ lies on a line passing through the origin and making equal angles with the axes,where $a$ is a constant. $A$ plane passes through $P$ and is perpendicular to $OP$. If this plane makes intercepts on the axes,what is the sum of the reciprocals of these intercepts?

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