Answer:
Option B
Explanation:
Vertices of $\triangle ABC$ are A(0,4,1) ,B(2,3,-1) and C(4,5,0)
AB= $\sqrt{(2-0)^{2}+(3-4)^{2}+(-1-1)^{2}}=\sqrt{4+1+4}=3$
$BC=\sqrt{(4-2)^{2}+(5-3)^{2}+(0+1)^{2}}=\sqrt{4+4+1}=3$
and
$CA=\sqrt{(4-0)^{2}+(5-4)^{2}+(0+1)^{2}}=\sqrt{16+1+4}=3\sqrt{2}$
$AB^{2}+BC^{2}=AC^{2}$
$\therefore$ $\triangle$ ABC is a right angled triangle
We know that , the orthocentre of a right angled triangle is the vertex containing the right angle
$\therefore$ Orthocentre is point B(2,3,-1)
