Vertices of $\Delta \mathrm{ABC}$ are $\mathrm{A}(0,4,1), \mathrm{B}(2,3,-1)$ and $\mathrm{C}(4,5,0)$.
$\begin{array}{l}
\begin{aligned}
\mathrm{AB} &=\sqrt{(2-0)^{2}+(3-4)^{2}+(-1-1)^{2}} \\
&=\sqrt{4+1+4}=3
\end{aligned} \\
\mathrm{BC}=\sqrt{(4-2)^{2}+(5-3)^{2}+(0+1)^{2}} \\
=\sqrt{4+4+1}=3 \\
\text { and } \mathrm{CA}=\sqrt{(4-0)^{2}+(5-4)^{2}+(0-1)^{2}} \\
\quad=\sqrt{16+1+1}=3 \sqrt{2}
\end{array}$
$\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}$
$\therefore \quad \Delta \mathrm{ABC}$ is a right angled triangle.
We know that, the orthocentre of a right angled triangle is the vertex containing the right angle.
$\therefore \quad$ Orthocentre is point $\mathrm{B}(2,3,-1)$.