A variable plane cut the coordinate axis at $x,y,z$ axis at point $A,B$ and $C$ respectively such that the volume of Tetrahedron $OABC$ is remain constant and equals $32$ cubic units and $O$ represent origin, Then
$(i)$ Locus of foot of perpendicar drawn from origin $O$ to plane $ABC$ is
$(ii)$ If $PA,PB,PC$ are mutually perpendicular, Then locus of point $P$ is
what i try
volume if Tetrahedron is $\displaystyle \frac{abc}{6}=32\cdots \cdots (1)$
Equation of plane $ABC$ is $\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Let locus of foot of perpendicular be $\displaystyle P(h,k,l)$
ut into $(1)$, get $\displaystyle \frac{h}{a}+\frac{k}{b}+\frac{l}{c}=1\cdots (2)$
and direction ratio of $OP$ is $\displaystyle\bigg<h,k,l\bigg>$
Then line $OP$ is perpendicular to plane $ABC$
and line perpendicar to that plane whose direction ratio is $\displaystyle \bigg<\frac{1}{a},\frac{1}{b},\frac{1}{c}\bigg>$
so $\displaystyle ha=hb=hc=\lambda$ put into $(2)$
$\displaystyle \lambda=\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}$
How do i solve it Help me please