A variable plane makes intercepts on x, y and z axes and it makes a tetrahedron of volume 64 cu. units The locus of foot of perpendicular from origin on this plane is

The equation of plane in the intercept form is 

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

Volume of tetrahedron $OABC\text{ is }v=\frac{abc}{6}=64\Rightarrow abc=384$

Foot of perpendicular from (0, 0, 0) on this plane is 

$\frac{x}{1/a}=\frac{y}{1/b}=\frac{z}{1/c}=\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=k$

$\Rightarrow \frac{1}{k}=\frac{x^2+y^2+z^2}{k^2}\Rightarrow x^2+y^2+z^2=k$

$\Rightarrow \frac{1}{k}=\frac{x^2+y^2+z^2}{k^2}\Rightarrow x^2+y^2+z^2=k$

$\therefore {\left(x^2+y^2+z^2\right)}^3=k^3=abcxyz=384xyz$ is the required  locus.