If A(4,3,2), B(5,4,6), C(-1,-1,5) are the ver | vedclass

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(A) The vertices of the triangle are $A(4,3,2), B(5,4,6)$, and $C(-1,-1,5)$.First, we calculate the lengths of sides $AB$ and $AC$ using the distance

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$\left(\frac{22}{8}, \frac{17}{8}, \frac{45}{8}\right)$

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(A) The vertices of the triangle are $A(4,3,2), B(5,4,6)$, and $C(-1,-1,5)$.First, we calculate the lengths of sides $AB$ and $AC$ using the distance formula:$AB = \sqrt{(5-4)^2 + (4-3)^2 + (6-2)^2} = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$.$AC = \sqrt{(-1-4)^2 + (-1-3)^2 + (5-2)^2} = \sqrt{(-5)^2 + (-4)^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}$.According to the Angle Bisector Theorem, the bisector of angle $A$ divides the opposite side $BC$ in the ratio of the sides containing the angle, which is $AB : AC = 3\sqrt{2} : 5\sqrt{2} = 3 : 5$.Let $D$ be the point on $BC$ where the bisector meets it. $D$ divides $BC$ internally in the ratio $m:n = 3:5$.Using the section formula $D = \left(\frac{mx_C + nx_B}{m+n}, \frac{my_C + ny_B}{m+n}, \frac{mz_C + nz_B}{m+n}\right)$:$x_D = \frac{3(-1) + 5(5)}{3+5} = \frac{-3 + 25}{8} = \frac{22}{8}$.$y_D = \frac{3(-1) + 5(4)}{3+5} = \frac{-3 + 20}{8} = \frac{17}{8}$.$z_D = \frac{3(5) + 5(6)}{3+5} = \frac{15 + 30}{8} = \frac{45}{8}$.Thus, the coordinates of point $D$ are $\left(\frac{22}{8}, \frac{17}{8}, \frac{45}{8}\right)$.

If $A(4,3,2), B(5,4,6), C(-1,-1,5)$ are the vertices of a triangle, then the coordinates of the point in which the bisector of the angle $A$ meets the side $BC$ is

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