Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 06 (e)/Solution 3

In this particular example there is an even faster way to see that $D$ must be outside of the triangle $T$, because $D$ is even outside the cuboid spanned by the three corners of $T$. Indeed, for all points $(x,y,z)$ in $T$ it must be true that

$\begin{array}{rl}\min (0,1,-1)& \le x\le \max (0,1,-1)\\ \min (1,1,2)& \le y\le \max (1,1,2)\\ \min (2,5,2)& \le z\le \max (2,5,2)\end{array}$

However, since the $y$ and $z$ coordinates of $D$ do not satisfy the inequalities above, $D$ is outside the cuboid, and hence outside of $T$.