Let A(4,3,5), B(1,-2,1), C(3,2,1) be the vert | vedclass

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(B) First, calculate the lengths of sides $AB$ and $AC$ using the distance formula:$AB = \sqrt{(4-1)^2 + (3-(-2))^2 + (5-1)^2} = \sqrt{3^2 + 5^2 + 4^2

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$\frac{3 \sqrt{5}}{4}$

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(B) First, calculate the lengths of sides $AB$ and $AC$ using the distance formula:$AB = \sqrt{(4-1)^2 + (3-(-2))^2 + (5-1)^2} = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$$AC = \sqrt{(4-3)^2 + (3-2)^2 + (5-1)^2} = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$According to the Angle Bisector Theorem, the internal bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the sides containing the angle:$\frac{BD}{DC} = \frac{AB}{AC} = \frac{5\sqrt{2}}{3\sqrt{2}} = \frac{5}{3}$Thus, point $D$ divides $BC$ internally in the ratio $5:3$. Using the section formula, the coordinates of $D$ are:$D = \left( \frac{5(3) + 3(1)}{5+3}, \frac{5(2) + 3(-2)}{5+3}, \frac{5(1) + 3(1)}{5+3} \right) = \left( \frac{18}{8}, \frac{4}{8}, \frac{8}{8} \right) = \left( \frac{9}{4}, \frac{1}{2}, 1 \right)$Now, calculate the length $CD$ using the distance formula between $C(3,2,1)$ and $D(\frac{9}{4}, \frac{1}{2}, 1)$:$CD = \sqrt{(\frac{9}{4} - 3)^2 + (\frac{1}{2} - 2)^2 + (1 - 1)^2}$$CD = \sqrt{(-\frac{3}{4})^2 + (-\frac{3}{2})^2 + 0^2} = \sqrt{\frac{9}{16} + \frac{9}{4}} = \sqrt{\frac{9 + 36}{16}} = \sqrt{\frac{45}{16}} = \frac{3\sqrt{5}}{4}$

Let $A(4,3,5), B(1,-2,1), C(3,2,1)$ be the vertices of a triangle $ABC$. If the internal bisector of $\angle BAC$ meets the side $BC$ at $D$, then $CD=$

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