A(3, 2, -1), B(4, 1, 0), C(2, 1, 4) are the v | vedclass

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(C) The lengths of the sides $AB$ and $AC$ are calculated as follows:$AB = \sqrt{(4-3)^2 + (1-2)^2 + (0-(-1))^2} = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3

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$3$

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(C) The lengths of the sides $AB$ and $AC$ are calculated as follows:$AB = \sqrt{(4-3)^2 + (1-2)^2 + (0-(-1))^2} = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$$AC = \sqrt{(2-3)^2 + (1-2)^2 + (4-(-1))^2} = \sqrt{(-1)^2 + (-1)^2 + 5^2} = \sqrt{1 + 1 + 25} = \sqrt{27} = 3\sqrt{3}$By the Angle Bisector Theorem, the bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the adjacent sides:$\frac{BD}{DC} = \frac{AB}{AC} = \frac{\sqrt{3}}{3\sqrt{3}} = \frac{1}{3}$Using the section formula, the coordinates of $D(p, q, r)$ are:$D = \left( \frac{1(2) + 3(4)}{1+3}, \frac{1(1) + 3(1)}{1+3}, \frac{1(4) + 3(0)}{1+3} \right) = \left( \frac{14}{4}, \frac{4}{4}, \frac{4}{4} \right) = \left( \frac{7}{2}, 1, 1 \right)$Thus, $p = \frac{7}{2}, q = 1, r = 1$.Finally, $\sqrt{2p + q + r} = \sqrt{2(\frac{7}{2}) + 1 + 1} = \sqrt{7 + 1 + 1} = \sqrt{9} = 3$.

$A(3, 2, -1), B(4, 1, 0), C(2, 1, 4)$ are the vertices of a triangle $ABC$. If the bisector of $\angle BAC$ intersects the side $BC$ at $D(p, q, r)$, then $\sqrt{2p + q + r} =$

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