Question

The position vectors of the four angular points of a tetrahedron $OABC$ are $(0,0,0);(0,0,2);(0,4,0)$ and $(6,0,0)$ respectively. A point $P$ inside the tetrahedron is at the same distance $r$ from the four plane faces of the tetrahedron. Find the value of $3r$.

Answer 1

Angular point $OABC$ are $(0,0,0),(0,0,2),(0,4,0)$ & $(6,0,0)$

Let centre of sphere be $(r,r,r)$

Equation of plane passing $ABC$ is

$\frac{x}{6}+\frac{y}{4}+\frac{z}{2}=1$

$r=\left|\frac{\frac{r}{6}+\frac{r}{4}+\frac{r}{2}-1}{\sqrt{\frac{1}{6^2}+\frac{1}{4^2}+\frac{1}{2^2}}}\right|$

$7r=\pm (11r-12)$

$r=\frac{2}{3}$, $r=3$ (not satisfied)