If A(2,-1,1),B(2,5,1),and C(0,-2,3) are the v | vedclass

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(D) The length of the sides of $	riangle ABC$ are calculated using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.$c = AB

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$\frac{4}{\sqrt{3}}$

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(D) The length of the sides of $	riangle ABC$ are calculated using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.$c = AB = \sqrt{(2-2)^2 + (5-(-1))^2 + (1-1)^2} = \sqrt{0^2 + 6^2 + 0^2} = 6$.$b = AC = \sqrt{(0-2)^2 + (-2-(-1))^2 + (3-1)^2} = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = 3$.$a = BC = \sqrt{(0-2)^2 + (-2-5)^2 + (3-1)^2} = \sqrt{(-2)^2 + (-7)^2 + 2^2} = \sqrt{4+49+4} = \sqrt{57}$.By the Angle Bisector Theorem,point $D$ divides $BC$ in the ratio $c:b$,which is $6:3 = 2:1$.Using the section formula,the coordinates of $D$ are $\left(\frac{2(0)+1(2)}{2+1}, \frac{2(-2)+1(5)}{2+1}, \frac{2(3)+1(1)}{2+1}ight) = \left(\frac{2}{3}, \frac{1}{3}, \frac{7}{3}ight)$.Now,$AD = \sqrt{(\frac{2}{3}-2)^2 + (\frac{1}{3}-(-1))^2 + (\frac{7}{3}-1)^2} = \sqrt{(-\frac{4}{3})^2 + (\frac{4}{3})^2 + (\frac{4}{3})^2} = \sqrt{\frac{16}{9} + \frac{16}{9} + \frac{16}{9}} = \sqrt{\frac{48}{9}} = \frac{4\sqrt{3}}{3} = \frac{4}{\sqrt{3}}$.

If $A(2,-1,1)$,$B(2,5,1)$,and $C(0,-2,3)$ are the vertices of a triangle,and $D$ is the point of intersection of the side $BC$ and the internal angular bisector of angle $A$,then $AD=$

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