The distance from a point (1, 1, 1) to a vari | vedclass

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(C) Let the equation of the plane $\pi$ be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $A = (a, 0, 0)$,$B = (0, b, 0)$,and $C = (0, 0, c)$.The

Vedclass · Accepted Answer

$\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - 1\right)^2 = 144\left(\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}\right)$

Vedclass · Answer

(C) Let the equation of the plane $\pi$ be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $A = (a, 0, 0)$,$B = (0, b, 0)$,and $C = (0, 0, c)$.The distance from point $(1, 1, 1)$ to the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} - 1 = 0$ is given by $d = \frac{|\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 12$.Squaring both sides,we get $\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1\right)^2 = 144\left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$.The point $P$ is the intersection of planes $x=a$,$y=b$,and $z=c$,so $P \equiv (a, b, c)$.Replacing $(a, b, c)$ with $(x, y, z)$,the locus of $P$ is $\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - 1\right)^2 = 144\left(\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}\right)$.

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