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

• A

  $\alpha x+\beta y+\gamma z=1$
• Correct answer
  B

  $\frac{\alpha}{x}+\frac{\beta}{y}+\frac{\gamma}{z}=1$
• C

  $\alpha x^2+\beta y^2+\gamma z^2=1$
• D

  $\alpha \beta x+\beta \gamma y+\alpha \gamma z=1$

Solution

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, $$ \frac{\alpha}{x}+\frac{\beta}{y}+\frac{\gamma}{z}=1 $$

Hence, the correct answer is option: 'B'.