If the mid-points of the sides AB, BC, CA of | vedclass

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(D) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.The mid-points of sides $AB, BC, CA$ are given as

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$5$

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(D) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.The mid-points of sides $AB, BC, CA$ are given as $D(1, 5, -1)$,$E(0, 4, -2)$,and $F(2, 3, 4)$ respectively.By the mid-point formula:For $AB$: $\frac{x_1+x_2}{2} = 1, \frac{y_1+y_2}{2} = 5, \frac{z_1+z_2}{2} = -1 \Rightarrow x_1+x_2 = 2, y_1+y_2 = 10, z_1+z_2 = -2$.For $BC$: $\frac{x_2+x_3}{2} = 0, \frac{y_2+y_3}{2} = 4, \frac{z_2+z_3}{2} = -2 \Rightarrow x_2+x_3 = 0, y_2+y_3 = 8, z_2+z_3 = -4$.For $CA$: $\frac{x_3+x_1}{2} = 2, \frac{y_3+y_1}{2} = 3, \frac{z_3+z_1}{2} = 4 \Rightarrow x_3+x_1 = 4, y_3+y_1 = 6, z_3+z_1 = 8$.Adding these equations:$2(x_1+x_2+x_3) = 2+0+4 = 6 \Rightarrow x_1+x_2+x_3 = 3$.$2(y_1+y_2+y_3) = 10+8+6 = 24 \Rightarrow y_1+y_2+y_3 = 12$.$2(z_1+z_2+z_3) = -2-4+8 = 2 \Rightarrow z_1+z_2+z_3 = 1$.Now,to find $C(x_3, y_3, z_3)$,subtract the equations for $AB$ from the sums:$x_3 = (x_1+x_2+x_3) - (x_1+x_2) = 3 - 2 = 1$.$y_3 = (y_1+y_2+y_3) - (y_1+y_2) = 12 - 10 = 2$.$z_3 = (z_1+z_2+z_3) - (z_1+z_2) = 1 - (-2) = 3$.So,$C = (1, 2, 3)$.The median from $C$ to $AB$ is the line segment $CD$,where $D$ is the mid-point of $AB$,$D(1, 5, -1)$.Length $CD = \sqrt{(1-1)^2 + (5-2)^2 + (-1-3)^2} = \sqrt{0^2 + 3^2 + (-4)^2} = \sqrt{0 + 9 + 16} = \sqrt{25} = 5$.

If the mid-points of the sides $AB, BC, CA$ of a triangle are $(1, 5, -1), (0, 4, -2), (2, 3, 4)$ respectively,then the length of the median drawn from $C$ to $AB$ is

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