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 :

Updated On: Oct 1, 2024
  • $\frac{x}{3} + \frac{y}{2} + \frac{z}{1} = 1 $
  • $ x + y + z = 6 $
  • $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{11}{6} $
  • $\frac{3}{x} + \frac{2}{y} + \frac{1}{z} = 1 $
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The Correct Option is D

Solution and Explanation

Let plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c} =1$
it passes through (3,2,1) $\therefore \frac{3}{a}+\frac{2}{b}+\frac{1}{c} = 1$
Now A (a,0,0), B (0, b, 0), C (0,0,c)
$\therefore$ Locus of point of intersection of planes x = a
y = b, z = c is $\frac{3}{x}+\frac{2}{y}+\frac{1}{z} = 1$
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