A plane passes through a fixed point (p, q, r | vedclass

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(A) Let the coordinates of $A, B,$ and $C$ be $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ respectively.The equation of the plane is $\frac{x}{a} + \frac{y

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$\frac{p}{x} + \frac{q}{y} + \frac{r}{z} = 2$

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(A) Let the coordinates of $A, B,$ and $C$ be $(a, 0, 0), (0, b, 0),$ and $(0, 0, c)$ 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 $(p, q, r)$,we have $\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$.The equation of the sphere passing through the origin $O(0, 0, 0)$ and points $A, B, C$ is given by $x^2 + y^2 + z^2 - ax - by - cz = 0$.The centre of this sphere is $(\frac{a}{2}, \frac{b}{2}, \frac{c}{2})$.Let the centre be $(x, y, z)$. Then $x = \frac{a}{2}, y = \frac{b}{2}, z = \frac{c}{2}$,which implies $a = 2x, b = 2y, c = 2z$.Substituting these values into the plane equation $\frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1$,we get $\frac{p}{2x} + \frac{q}{2y} + \frac{r}{2z} = 1$.Multiplying by $2$,the locus of the centre is $\frac{p}{x} + \frac{q}{y} + \frac{r}{z} = 2$.

$A$ plane passes through a fixed point $(p, q, r)$ and cuts the axes at $A, B, C$. Then the locus of the centre of the sphere $OABC$ is

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