A variable plane passes through a fixed point | vedclass

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(C) Let the intercepts of the variable plane on the $x, y,$ and $z$ axes be $a, b,$ and $c$ respectively.The equation of the plane is $\frac{x}{a} + \

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the intercepts of the variable plane on the $x, y,$ and $z$ axes be $a, b,$ and $c$ respectively.The equation of the plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Since the plane passes through the fixed point $(3, 2, 1)$,we have $\frac{3}{a} + \frac{2}{b} + \frac{1}{c} = 1$.The plane through $A(a, 0, 0)$ parallel to the $yz$-plane is $x = a$.The plane through $B(0, b, 0)$ parallel to the $zx$-plane is $y = b$.The plane through $C(0, 0, c)$ parallel to the $xy$-plane is $z = c$.The point of intersection of these three planes is $(a, b, c)$.Let the point of intersection be $(x, y, z)$. Then $x = a, y = b,$ and $z = c$.Substituting these into the equation $\frac{3}{a} + \frac{2}{b} + \frac{1}{c} = 1$,we get the locus $\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1$.

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