Answer:
[tex]5/2units\\[/tex].
[tex]\sqrt{47/2}units \\[/tex].
[tex]\sqrt{41/2}units\\[/tex].
Step-by-step explanation:
The given points are A(1,2,3), B(-2,0,5) and C(4,1,5). The triangle is represented in the attach file where the three possible median are length AE, BF, and CD. We determine the coordinate of point D,E and F using the midpoint equation which is for any point A(x,y,z) and point B(a,b,c), the midpoint D is determine by
[tex]D=(\frac{x+a}{2},\frac{y+b}{2},\frac{z+c}{2})\\[/tex].
Hence going by the above formula we determine the coordinate of point D,E and F
[tex]D=(\frac{-2+1}{2},\frac{0+2}{2},\frac{5+3}{2})\\[/tex].
[tex]D=(\frac{-1}{2},1,4)\\[/tex].
point E
[tex]E=(\frac{4-2}{2},\frac{1+0}{2},\frac{5+5}{2})\\[/tex].
[tex]E=(1,\frac{1}{2},5)\\[/tex].
Point F
[tex]F=(\frac{4+1}{2},\frac{1+2}{2},\frac{5+3}{2})\\[/tex].
[tex]F=(\frac{5}{2},\frac{3}{2},4)\\[/tex].
To determine the length of each median line we use the formula for distance between two points which is express as
[tex]AB=\sqrt{(y_{2}-y_{1} )^{2} +(x_{2}-x_{1} )^{2}+(z_{2}-z_{1} )^{2}} \\[/tex].
Using the above formula we determine the length of line AE,BF and CD.
[tex]AE=\sqrt{(1-1 )^{2} +(1/2-2)^{2}+(5-3 )^{2}} \\[/tex].
[tex]AE=\sqrt{0 +9/4+4} \\[/tex].
[tex]AE=\sqrt{25/4} \\[/tex].
[tex]AE=5/2units\\[/tex].
For point BF
[tex]BF=\sqrt{(5/2+2 )^{2} +(3/2-0)^{2}+(4-5 )^{2}}\\[/tex].
[tex]BF=\sqrt{81/4 +9/4+1} \\[/tex].
[tex]BF=\sqrt{47/2} \\[/tex].
[tex]BF=\sqrt{47/2}units \\[/tex].
For point CD
[tex]CD=\sqrt{(-1/2-4 )^{2} +(1-1)^{2}+(4-5 )^{2}}\\[/tex].
[tex]BF=\sqrt{81/4 +0+1} \\[/tex].
[tex]BF=\sqrt{41/2} \\[/tex].
[tex]BF=\sqrt{41/2}units\\[/tex].
