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A variable plane passes through a fixed point $(\alpha, \beta, \gamma)$ and meets the coordinate axes in $A, B$ and $C$. Let $P_1, P_2$ and $P_3$ be the planes passing through $A, B, C$ and parallel to the coordinate planes $Y Z, Z X, X Y$ respectively. Then, the locus of the point of intersection of the planes $P_1, P_2$ and $P_3$ is
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Verified Answer
The correct answer is:
$\frac{\alpha}{x}+\frac{\beta}{y}+\frac{\gamma}{z}=1$
Let the point $A(a, 0,0), B(0, b, 0)$ and $C(0,0, c)$.
So, point of intersection of planes $P_1, P_2$ and $P_3$ is $P(a, b, c)$
Now, equation of plane $A B C$ is

$\because$ plane Eq. (i) passes through the point $(\alpha, \beta, \gamma)$, so
$$
\frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}=1
$$
On taking locus of point $(a, b, c)$ we are getting,
$$
\frac{\alpha}{x}+\frac{\beta}{y}+\frac{\gamma}{z}=1
$$
So, point of intersection of planes $P_1, P_2$ and $P_3$ is $P(a, b, c)$
Now, equation of plane $A B C$ is

$\because$ plane Eq. (i) passes through the point $(\alpha, \beta, \gamma)$, so
$$
\frac{\alpha}{a}+\frac{\beta}{b}+\frac{\gamma}{c}=1
$$
On taking locus of point $(a, b, c)$ we are getting,
$$
\frac{\alpha}{x}+\frac{\beta}{y}+\frac{\gamma}{z}=1
$$
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