A triangle has vertices at the points A(5,0,0), B(0,5,0), and C(0,0,5). Identify the point in the interior of the triangle that is closest to the origin.
1. The points A, B, C define a plane:
$\displaystyle px,y,z)=(5,0,0)+r(5,-5,0+s(5,0,-5)$
2. p in normal form:
$\displaystyle \vec n = (5,-5,0) \times (5,0,-5) = (5,5,5)$ Thus
$\displaystyle p: x+y+z=5$
3. The point F lies on the line through the origin perpendicular to p:
$\displaystyle (x,y,z)=r (1,1,1)$
4. Calculate the point of intersection:
$\displaystyle r+r+r = 5~\implies~r=\dfrac53$
Therefore $\displaystyle F\left(\dfrac53\ ,\ \dfrac53\ ,\ \dfrac53\right)$