Show that the triangle ABC with vertices A (0, 4, 1),

Given vertices are $A(0,4,1), B(2,3,-1)$ and $C(4,5,0)$.

To prove right angled triangle, consider

$\mathrm{AB}=\sqrt{(0-2)^{2}+(4-3)^{2}+(1+1)^{2}}=\sqrt{4+1+4}=\sqrt{9}=3$

$B C=\sqrt{(2-4)^{2}+(3-5)^{2}+(-1-0)^{2}}=\sqrt{4+4+1}=\sqrt{9}=3$

$\mathrm{AC}=\sqrt{(0-4)^{2}+(4-5)^{2}+(1-0)^{2}}=\sqrt{16+1+1}=\sqrt{18}=3 \sqrt{2}$

$\Rightarrow \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$

$\therefore$ Triangle $\mathrm{ABC}$ a right angled.