Show that the points $A(1, 2, 3)$,$B(-1, -2, -1)$,$C(2, 3, 2)$,and $D(4, 7, 6)$ are the vertices of a parallelogram $ABCD$,but it is not a rectangle.

To show that $ABCD$ is a parallelogram,we need to show that opposite sides are equal.
$AB = \sqrt{(-1-1)^{2} + (-2-2)^{2} + (-1-3)^{2}} = \sqrt{(-2)^{2} + (-4)^{2} + (-4)^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$BC = \sqrt{(2 - (-1))^{2} + (3 - (-2))^{2} + (2 - (-1))^{2}} = \sqrt{3^{2} + 5^{2} + 3^{2}} = \sqrt{9 + 25 + 9} = \sqrt{43}$
$CD = \sqrt{(4-2)^{2} + (7-3)^{2} + (6-2)^{2}} = \sqrt{2^{2} + 4^{2} + 4^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$DA = \sqrt{(1-4)^{2} + (2-7)^{2} + (3-6)^{2}} = \sqrt{(-3)^{2} + (-5)^{2} + (-3)^{2}} = \sqrt{9 + 25 + 9} = \sqrt{43}$
Since $AB = CD$ and $BC = DA$,$ABCD$ is a parallelogram.
Now,to prove that $ABCD$ is not a rectangle,we show that the diagonals $AC$ and $BD$ are unequal.
$AC = \sqrt{(2-1)^{2} + (3-2)^{2} + (2-3)^{2}} = \sqrt{1^{2} + 1^{2} + (-1)^{2}} = \sqrt{1 + 1 + 1} = \sqrt{3}$
$BD = \sqrt{(4 - (-1))^{2} + (7 - (-2))^{2} + (6 - (-1))^{2}} = \sqrt{5^{2} + 9^{2} + 7^{2}} = \sqrt{25 + 81 + 49} = \sqrt{155}$
Since $AC \neq BD$,$ABCD$ is not a rectangle.