Question

The position vectors of the four angular points of a tetrahedron $OABC$ are $(0,0,0),(0,0,2),(0,4,0),(6,0,0)$ respectively. A point $P$ inside the tetrahedron is at the same distance $r$ from the four plane faces of the tetrahedron. Then the value of $9r$ is

Answer 1

The three faces of tetrahedron are $xy,yz,zx$ plane.
Since, point $P$ is equidistance from $xy,yz,zx$ plane, then its coordinates are $P(r,r,r)$.
Equation of the plane $ABC$ is,
$\frac{x}{6}+\frac{y}{4}+\frac{z}{2}=1$
$\Rightarrow 2x+3y+6z-12=0$
$P$ is also at a distance of $r$ from the plane $ABC$.
$\therefore \frac{|2r+3r+6r-12|}{\sqrt{2^2+3^3+6^2}}=r$
Squaring both the sides,
$\frac{(11r-12{)}^2}{49}=r^2\Rightarrow 9r^2-33r+18=0$
$\Rightarrow r=\frac{2}{3},3$
Since, point $P(r,r,r)$ lies inside the tetrahedron,
$\therefore r<2\Rightarrow r=\frac{2}{3}$