Question

A variable plane cuts the $x$-axis, $y$-axis and $z$-axis at the points $A,B$ and $C$ respectively such that the volume of the tetrahedron $OABC$ remains constant equal to $32$ cubic unit and $O$ is the origin of the coordinate system

Answer 1

Volume of tetrahedron $OABC$ $=\frac{1}{6}\left|\begin{array}{ccc}a& 0& 0\\ 0& b& 0\\ 0& 0& c\end{array}\right|=\frac{1}{6}abc=32$

A) Let centroid be $(h,k,l)$ then $h=\frac{a}{4},k=\frac{b}{4},l=\frac{c}{4}$
Gives $\frac{abc}{64}=hkl$
$\Rightarrow abc=64hkl\Rightarrow 32\times 6=64hkl$
$\Rightarrow hkl=3$
Hence $xyz=3$
B) Let point equidistant from $O,A,B$ and $C$ be $(h,k,l)$
Then $h=\frac{a}{2},k=\frac{b}{2},l=\frac{c}{2}$
$\frac{abc}{8}=hkl$
$\Rightarrow abc=8hkl$
$\Rightarrow 32\times 6=8hkl$
$\Rightarrow xyz=24$
C) Let the equation of plane be

$h(x-h)+k(y-k)+l(z-l)=0\Rightarrow hx+ky+lz=h^2+k^2+l^2$

Now as $A$ satisfies the equation plane then
$ah=h^2+k^2+l^2\Rightarrow a=\frac{h^2+k^2+l^2}{h}$

Similarly $b=\frac{h^2+k^2+l^2}{k}$ and $c=\frac{h^2+k^2+l^2}{l}$

Then $abc=\frac{(h^2+k^2+l^2{)}^3}{hkl}$

Therefore locus is $(x^2+y^2+z^2{)}^3=192xyz$
D) Let the point be $(h,k,l)$, then
$\overrightarrow{PA}=(a-h)\hat{i}-k\hat{j}-l\hat{k}$
$\overrightarrow{PB}=h\hat{i}+(b-k)\hat{j}-l\hat{k}$
$\overrightarrow{PC}=-h\hat{i}-k\hat{j}-(c-l)\hat{k}$

Now
$\overrightarrow{PA}\cdot \overrightarrow{PB}=0\Rightarrow -h(a-h)-k(b-k)+l^2=0\Rightarrow h^2+k^2+l^2=ah+bk$ ...(1)

Similarly $h^2+k^2+l^2=bk+cl$ ...(2)

And $h^2+k^2+l^2=cl+ah$ ...(3)

From (1), (2) and (3) we get $ah=bk=cl$
Therefore,
$a=\frac{h^2+k^2+l^2}{2h},b=\frac{h^2+k^2+l^2}{2k},c=\frac{h^2+k^2+l^2}{2l}$

$abc=\frac{(h^2+k^2+l^2{)}^3}{8hkl}$
$\Rightarrow 32\times 6\times 8xyz=(x^2+y^2+z^2{)}^3$
$\Rightarrow (x^2+y^2+z^2{)}^3=1536xyz$