Verify that the points $A(-1, 2, 1)$,$B(1, -2, 5)$,$C(4, -7, 8)$,and $D(2, -3, 4)$ are the vertices of a parallelogram.

(N/A) Let the given points be $A(-1, 2, 1)$,$B(1, -2, 5)$,$C(4, -7, 8)$,and $D(2, -3, 4)$.
First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:
$AB = \sqrt{(1 - (-1))^2 + (-2 - 2)^2 + (5 - 1)^2} = \sqrt{2^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$BC = \sqrt{(4 - 1)^2 + (-7 - (-2))^2 + (8 - 5)^2} = \sqrt{3^2 + (-5)^2 + 3^2} = \sqrt{9 + 25 + 9} = \sqrt{43}$
$CD = \sqrt{(2 - 4)^2 + (-3 - (-7))^2 + (4 - 8)^2} = \sqrt{(-2)^2 + 4^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$DA = \sqrt{(-1 - 2)^2 + (2 - (-3))^2 + (1 - 4)^2} = \sqrt{(-3)^2 + 5^2 + (-3)^2} = \sqrt{9 + 25 + 9} = \sqrt{43}$
Since $AB = CD = 6$ and $BC = DA = \sqrt{43}$,the opposite sides are equal.
Additionally,we check the diagonals $AC$ and $BD$:
$AC = \sqrt{(4 - (-1))^2 + (-7 - 2)^2 + (8 - 1)^2} = \sqrt{5^2 + (-9)^2 + 7^2} = \sqrt{25 + 81 + 49} = \sqrt{155}$
$BD = \sqrt{(2 - 1)^2 + (-3 - (-2))^2 + (4 - 5)^2} = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$
Since the opposite sides are equal and the diagonals are not equal,the quadrilateral $ABCD$ is a parallelogram.