From a moving point ‘P’ lying on the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = p$ . Perpendiculars PA, PB, PC are drawn to coordinate planes. The locus of the foot of perpendicular drawn from origin to plane passing through points A, B and C is.

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$(x^{2} + y^{2} + z^{2}) (\frac{1}{a x} + \frac{1}{b y} + \frac{1}{c z}) = p$

$(x^{2} + y^{2} + z^{2}) (\frac{a}{x} + \frac{b}{y} + \frac{c}{z}) = p$

$(x^{2} + y^{2} + z^{2}) (\frac{1}{a x} + \frac{1}{b y} + \frac{1}{c z}) = 2 p$

$(\frac{1}{x^{2}} + \frac{1}{y^{2}} + \frac{1}{z^{2}}) (\frac{x}{a} + \frac{y}{b} + \frac{z}{c}) = p$

answer is B.

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Detailed Solution

Let p be $(α, β, γ)$ so points A, B and C are $(α, β, 0), (α, 0, γ)$ and $(0, β, γ)$ respectively. So, plane passing through these points is $\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 2$ also $\frac{α}{a} + \frac{β}{b} + \frac{γ}{c} = p$ . If $(x_{1}, y_{1}, z_{1})$ is foot of perpendicular from origin, then equation of pane $x_{1} (x - x_{1}) + y_{1} (y - y_{1}) + z_{1} (z - z_{1}) = 0$
Comparing with $\frac{x}{α} + \frac{y}{β} + \frac{z}{γ} = 2$ , we get $2 α x_{1} = 2 β y_{1} = 2 γ z_{1} = x_{1}^{2} + y_{1}^{2} + z_{1}^{2}$
So locus is $(x_{1}^{2} + y_{1}^{2} + z_{1}^{2}) (\frac{1}{a x_{1}} + \frac{1}{b y_{1}} + \frac{1}{c z_{1}}) = 2 p$