Question

A variable plane passes through a fixed point $(a,b,c)$ and meets coordinate axes in $A,B,C$. The locus of the point common to the plane through $A,B,C$ parallel to coordinate planes is

• A. $ayz+bzx+cxy=xyz$
• B. $ayz+bzx+cxy=2xyz$
• C. $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1$
• D. none of these

Answer 1

Answer: (A), (C)

The correct options are
A $ayz+bzx+cxy=xyz$
C $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1$

Let the plane be $\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$

It passes through $(a,b,c),\therefore \frac{a}{\alpha }+\frac{b}{\beta }+\frac{c}{\gamma }=1$ ...$\left(1\right)$

Now coordinates of the points $A,B,C$ are $(\alpha ,0,0),(0,\beta ,0)$ and $(0,0,\gamma )$ respectively.

Equation of the planes through $A,B,C$ parallel to coordinate planes are

$x=\alpha$ ...$\left(2\right)$

$y=\beta$ ...$\left(3\right)$

and $z=\gamma$ ...$\left(4\right)$

The locus of their point of intersection will be obtained by elimination $\alpha ,\beta ,\gamma$ from these with the help of the relation $\left(1\right)$.

We thus get $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1$

$\Rightarrow ayz+bxz+cxy=xyz$ which is the required locus.