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

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

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