Question

A variable plane passes through a fixed point $(3,2,1)$ and meets $x,y$ and $z$ axes at $A,B$ and $C$ respectively. A plane is drawn parallel to $yz-$plane through $A$, a second plane is drawn parallel $zx-$plane through $B$ and a third plane is drawn parallel to $xy-$plane through $C$. Then the locus of the point of intersection of these three planes, is :

• A. $\frac{x}{3}+\frac{y}{2}+\frac{z}{1}=1$
• B. $x+y+z=6$
• C. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{11}{6}$
• D. $\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1$

Answer 1

The correct option is D $\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1$

Let the equation of the plane be,
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
So the points,
$A=(a,0,0)B=(0,b,0)C=(0,0,c)$
We know that the plane passes through $(3,2,1)$
So,
$\frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1\cdots \left(1\right)$
Now,
Plane parallel to $yz-$ plane and paasing through $A$,
$x=a$,
Similarly the other planes will be,
$y=b,z=c$
Point of intersection will be ,
$(a,b,c)$
From equation $\left(1\right),$ the locus of point of intersection will be,
$\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1$