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(N/A) Let the vertices of the triangle be $A(x_1, y_1, z_1), B(x_2, y_2, z_2)$ and $C(x_3, y_3, z_3)$.Let the mid-points be $D(1, 5, -1)$ on $BC$,$E(0

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(N/A) Let the vertices of the triangle be $A(x_1, y_1, z_1), B(x_2, y_2, z_2)$ and $C(x_3, y_3, z_3)$.
Let the mid-points be $D(1, 5, -1)$ on $BC$,$E(0, 4, -2)$ on $AC$,and $F(2, 3, 4)$ on $AB$.
Using the mid-point formula:
$\frac{x_1+x_2}{2} = 2, \frac{y_1+y_2}{2} = 3, \frac{z_1+z_2}{2} = 4 \implies x_1+x_2 = 4, y_1+y_2 = 6, z_1+z_2 = 8$ $(i)$
$\frac{x_1+x_3}{2} = 0, \frac{y_1+y_3}{2} = 4, \frac{z_1+z_3}{2} = -2 \implies x_1+x_3 = 0, y_1+y_3 = 8, z_1+z_3 = -4$ (ii)
$\frac{x_2+x_3}{2} = 1, \frac{y_2+y_3}{2} = 5, \frac{z_2+z_3}{2} = -1 \implies x_2+x_3 = 2, y_2+y_3 = 10, z_2+z_3 = -2$ (iii)
Adding $(i)$,(ii),and (iii): $2(x_1+x_2+x_3) = 6 \implies x_1+x_2+x_3 = 3$. Similarly,$y_1+y_2+y_3 = 12$ and $z_1+z_2+z_3 = 1$.
Subtracting (iii) from the sum: $x_1 = 3-2 = 1, y_1 = 12-10 = 2, z_1 = 1-(-2) = 3$. So,$A = (1, 2, 3)$.
Subtracting (ii) from the sum: $x_2 = 3-0 = 3, y_2 = 12-8 = 4, z_2 = 1-(-4) = 5$. So,$B = (3, 4, 5)$.
Subtracting $(i)$ from the sum: $x_3 = 3-4 = -1, y_3 = 12-6 = 6, z_3 = 1-8 = -7$. So,$C = (-1, 6, -7)$.
Centroid $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right) = \left(\frac{3}{3}, \frac{12}{3}, \frac{1}{3}\right) = (1, 4, 1/3)$.

The mid-points of the sides of a triangle are $(1, 5, -1), (0, 4, -2)$ and $(2, 3, 4)$. Find its vertices. Also,find the centroid of the triangle.

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