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

Answer 1

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

Let a plane be,

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

It passes through (3,2,1).

$\therefore \frac{3}{a}+\frac{2}{b}+\frac{1}{c}=1......\left(1\right)$

Now, coordinates of points A, B, C are (a, 0, 0), (0, b, 0), (0, 0, c) respectively.

Equation of the plane through A, B, C parellel to coordinate plane are

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

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

and

$z=c.......\left(4\right)$

The locus of their point of intersection will be obtained by eliminating a, b, c from these with the help of eq(1).

Thus, we get

$\frac{3}{x}+\frac{2}{y}+\frac{1}{z}=1$