A variable plane makes intercepts on X, Y and Z-axes and it makes a tetrahedron of volume 64cu. Units. The locus of foot of perpendicular from origin on this plane is

Options

A. `(x^(2)+y^(2)+z^(2))^(2)=384 xyz`
B. `xyz=681`
C. `(x+y+z)(1/x+1/y+1/z)^(2)=16`
D. `xyz(x+y+z)=81`

Answer

A

Explanation

The equation of plane in the intercept form is `x/a+y/b+z/c=1`.
Volume of tetrahedron OABC is
`v=(abc)/6=64 rArr abc=384`
Foot of perpendicular forrm (0,0,0) on this plane is
`x/(1//a)=y/(1//b)=z/(1//c)=1/(1/a^(2) +1/b^(2)+1/c^(2))`=k
`rArr x=k/a, y=k/b, z=k/c`
and `1/k=1/a^(2)+1/b^(2)+1/c^(2)`
`rArr 1/k=(x^(2)+y^(2)+z^(2))/k^(2) rArr x^(2)+y^(2)+z^(2)=`k
`therefore (x^(2)+y^(2)+z^(2))^(2)=abc xyz=384 xyz` is the required locus.