A variable plane cuts off intercepts from the co-ordinate axes which are equal to the roots of the equation $x^{3} + 5 x = p - q x^{2} (p, q are real numbers) .$ The locus of the foot of the perpendicular from the origin to the plane is

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${(x^{2} + y^{2} + z^{2})}^{2} (x y + y z + z x) = 5$

${(x^{2} + y^{2} + z^{2})}^{4} (\frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x}) = 5$

${(x^{2} + y^{2} + z^{2})}^{2} (\frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x}) = 5$

$(x^{2} + y^{2} + z^{2}) (x y + y z + z x) = 5$

answer is C.

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Detailed Solution

Let a,b,c be the intercepts cut off by the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ or $l x + m y + n z = p$ on the coordinate axis.
We know $a = \frac{p}{l}, b = \frac{p}{m}, c = \frac{p}{n}$ where p is plane and l,m,n are the direction cosines of the normals.
Foot of perpendicular from origin to the plane is $(α, β, γ) = (p l, p m, p n)$
Now, $a b + b c + c a = 5$
$\Rightarrow p^{4} (\frac{1}{α β} + \frac{1}{β γ} + \frac{1}{γ α}) = 5 \Rightarrow {(α^{2} + β^{2} + γ^{2})}^{2} (\frac{1}{α β} + \frac{1}{β γ} + \frac{1}{γ α}) = 5$