P is a fixed point (a, a, a) on a line throug | vedclass

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(D) The line passes through the origin $(0, 0, 0)$ and is equally inclined to the axes,so its direction ratios are $(1, 1, 1)$.The point $P$ is $(a, a

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$\frac{1}{a}$

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(D) The line passes through the origin $(0, 0, 0)$ and is equally inclined to the axes,so its direction ratios are $(1, 1, 1)$.The point $P$ is $(a, a, a)$. The vector $\vec{OP}$ is the normal to the plane,so the normal vector is $\vec{n} = a\hat{i} + a\hat{j} + a\hat{k}$,which is parallel to $\hat{i} + \hat{j} + \hat{k}$.The equation of the plane passing through $(a, a, a)$ with normal vector $(1, 1, 1)$ is:$1(x - a) + 1(y - a) + 1(z - a) = 0$$x + y + z = 3a$To find the intercepts,we write the equation in the intercept form $\frac{x}{X} + \frac{y}{Y} + \frac{z}{Z} = 1$:$\frac{x}{3a} + \frac{y}{3a} + \frac{z}{3a} = 1$The intercepts on the axes are $X = 3a$,$Y = 3a$,and $Z = 3a$.The sum of the reciprocals of the intercepts is:$\frac{1}{3a} + \frac{1}{3a} + \frac{1}{3a} = \frac{3}{3a} = \frac{1}{a}$.

$P$ 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:

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