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(C) Let the equation of the variable plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since it passes through the fixed point $(3, 2, 1)$,we ha

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$\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1$

Vedclass · Answer

(C) Let the equation of the variable plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Since it passes through the fixed point $(3, 2, 1)$,we have $\frac{3}{a} + \frac{2}{b} + \frac{1}{c} = 1$ (Equation $i$). The coordinates of points $A, B,$ and $C$ are $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ respectively. The plane passing through $A(a, 0, 0)$ parallel to the $YZ$-plane is $x = a$. The plane passing through $B(0, b, 0)$ parallel to the $ZX$-plane is $y = b$. The plane passing through $C(0, 0, c)$ parallel to the $XY$-plane is $z = c$. The point of intersection of these three planes is $(x, y, z) = (a, b, c)$. Substituting $a = x, b = y,$ and $c = z$ into Equation $i$,we get the locus: $\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1$.

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