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The position vectors of the four angular...

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 `r`

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The correct Answer is:
`6`
The given points are `O(0, 0, 0), A(0, 0, 2), B(0, 4, 0) and C(6, 0, 0)`
Here three faces of tetrahedron are `xy, yz, zx` plane.
Since point P is equidistance from `zx, xy and yz` planes, its coordinates are `P(r, r, r)`
Equation of plane `ABC` is
`" "2x+3y+6z=12` (from intercept form)
P is also at distance r from plane `ABC`. Thus,
`" "(|2r+3r+6r-12|)/(sqrt(4+9+36))= r`
or `" "|11r -12|= 7r`
or `" "11r-12= pm 7r`
or `" "r= (12)/(18), 3`
`therefore" "r= 2//3` (as `rlt2`)
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