If A(0,1,2),B(2,-1,3),and C(1,-3,1) are the v | vedclass

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(A) First,calculate the side lengths of $	riangle ABC$: $AB^2 = (2-0)^2 + (-1-1)^2 + (3-2)^2 = 4 + 4 + 1 = 9 \implies AB = 3$. $BC^2 = (1-2)^2 + (-3+

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$\frac{3}{\sqrt{2}}$

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(A) First,calculate the side lengths of $	riangle ABC$: $AB^2 = (2-0)^2 + (-1-1)^2 + (3-2)^2 = 4 + 4 + 1 = 9 \implies AB = 3$. $BC^2 = (1-2)^2 + (-3+1)^2 + (1-3)^2 = 1 + 4 + 4 = 9 \implies BC = 3$. $AC^2 = (1-0)^2 + (-3-1)^2 + (1-2)^2 = 1 + 16 + 1 = 18 \implies AC = 3\sqrt{2}$. Since $AB^2 + BC^2 = 9 + 9 = 18 = AC^2$,the triangle is a right-angled triangle with the right angle at $B$. In a right-angled triangle,the orthocentre $(H)$ is the vertex containing the right angle,so $H = B(2,-1,3)$. The circumcentre $(O)$ is the midpoint of the hypotenuse $AC$. $O = \left( \frac{0+1}{2}, \frac{1-3}{2}, \frac{2+1}{2} ight) = \left( \frac{1}{2}, -1, \frac{3}{2} ight)$. The distance $OH$ is given by $\sqrt{(2 - 1/2)^2 + (-1 - (-1))^2 + (3 - 3/2)^2}$. $OH = \sqrt{(3/2)^2 + 0^2 + (3/2)^2} = \sqrt{9/4 + 9/4} = \sqrt{18/4} = \frac{3\sqrt{2}}{2} = \frac{3}{\sqrt{2}}$.

If $A(0,1,2)$,$B(2,-1,3)$,and $C(1,-3,1)$ are the vertices of a triangle,then the distance between its circumcentre and orthocentre is

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