Solution:
The given vertices are $A(0,4,1), B(2,3,-1)$ and $C(4,5,0)$
We need to prove right angled triangle, consider
$\begin{array}{l}
\mathrm{AB}=\sqrt{(0-2)^{2}+(4-3)^{2}+(1+1)^{2}}=\sqrt{4+1+4}=\sqrt{9}=3 \\
\mathrm{BC}=\sqrt{(2-4)^{2}+(3-5)^{2}+(-1-0)^{2}}=\sqrt{4+4+1}=\sqrt{9}=3 \\
\mathrm{AC}=\sqrt{(0-4)^{2}+(4-5)^{2}+(1-0)^{2}}=\sqrt{16+1+1}=\sqrt{18}=3 \sqrt{2} \\
\Rightarrow \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}
\end{array}$
$\therefore$ The triangle $\mathrm{ABC}$ a right angled.