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

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(B) First,calculate the lengths of the sides of $	riangle ABC$:$AB = \sqrt{(2-1)^2 + (0-2)^2 + (1-0)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$$AC = \sqrt{(-3

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$\frac{2 \sqrt{14}}{3}$

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(B) First,calculate the lengths of the sides of $	riangle ABC$:$AB = \sqrt{(2-1)^2 + (0-2)^2 + (1-0)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$$AC = \sqrt{(-3-1)^2 + (0-2)^2 + (2-0)^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6}$$BC = \sqrt{(-3-2)^2 + (0-0)^2 + (2-1)^2} = \sqrt{25 + 0 + 1} = \sqrt{26}$Since $AD$ is the internal angle bisector of $\angle BAC$,it divides the opposite side $BC$ in the ratio of the adjacent sides $AB:AC = \sqrt{6}:2\sqrt{6} = 1:2$.Using the section formula,the coordinates of point $D$ are:$D = \left( \frac{1(-3) + 2(2)}{1+2}, \frac{1(0) + 2(0)}{1+2}, \frac{1(2) + 2(1)}{1+2} ight) = \left( \frac{1}{3}, 0, \frac{4}{3} ight)$The length of $AD$ is:$AD = \sqrt{(1 - 1/3)^2 + (2 - 0)^2 + (0 - 4/3)^2} = \sqrt{(2/3)^2 + 2^2 + (-4/3)^2} = \sqrt{4/9 + 4 + 16/9} = \sqrt{20/9 + 36/9} = \sqrt{56/9} = \frac{\sqrt{56}}{3} = \frac{2\sqrt{14}}{3}$

If $A(1, 2, 0)$,$B(2, 0, 1)$,and $C(-3, 0, 2)$ are the vertices of $\triangle ABC$,then the length of the internal bisector of $\angle BAC$ is

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