Show that the points A

To prove: Points A, B, C, D form parallelogram

Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by

$\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$

Here

$\left(x_{1}, y_{1}, z_{1}\right)=(1,2,3)$

$\left(x_{2}, y_{2}, z_{2}\right)=(-1,-2,-1)$

$\left(x_{3}, y_{3}, z_{3}\right)=(2,3,2)$

$\left(x_{4}, y_{4}, z_{4}\right)=(4,7,6)$

Length $A B=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

$=\sqrt{(-1-1)^{2}+(-2-2)^{2}+(-1-3)^{2}}$

$=\sqrt{(-2)^{2}+(-4)^{2}+(-4)^{2}}$

$=\sqrt{4+16+16}$

$=\sqrt{36}$

Length $B C=\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}+\left(z_{3}-z_{2}\right)^{2}}$

$=\sqrt{(2+1)^{2}+(3+2)^{2}+(2+1)^{2}}$

$=\sqrt{(3)^{2}+(5)^{2}+(3)^{2}}$

$=\sqrt{9+25+9}$

$=\sqrt{43}$

Length $C D=\sqrt{\left(x_{4}-x_{3}\right)^{2}+\left(y_{4}-y_{3}\right)^{2}+\left(z_{4}-z_{3}\right)^{2}}$

$=\sqrt{(4-2)^{2}+(7-3)^{2}+(6-2)^{2}}$

$=\sqrt{(2)^{2}+(4)^{2}+(4)^{2}}$

$=\sqrt{4+16+16}$

$=\sqrt{36}$

Length $A D=\sqrt{\left(x_{4}-x_{1}\right)^{2}+\left(y_{4}-y_{1}\right)^{2}+\left(z_{4}-z_{1}\right)^{2}}$

$=\sqrt{(4-1)^{2}+(7-2)^{2}+(6-3)^{2}}$

$=\sqrt{(3)^{2}+(5)^{2}+(3)^{2}}$

$=\sqrt{9+25+9}$

$=\sqrt{43}$

Length $A C=\sqrt{\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}+\left(z_{3}-z_{1}\right)^{2}}$

$=\sqrt{(2-1)^{2}+(3-2)^{2}+(2-3)^{2}}$

$=\sqrt{(1)^{2}+(1)^{2}+(-1)^{2}}$

$=\sqrt{1+1+1}$

$=\sqrt{3}$

Length $\mathrm{BD}=\sqrt{\left(\mathrm{x}_{4}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{4}-\mathrm{y}_{2}\right)^{2}+\left(\mathrm{z}_{4}-\mathrm{z}_{2}\right)^{2}}$

$=\sqrt{(4+1)^{2}+(7+2)^{2}+(6+1)^{2}}$

$=\sqrt{(5)^{2}+(9)^{2}+(7)^{2}}$

$=\sqrt{25+81+49}$

$=\sqrt{155}$

Here, AB = CD which are opposite sides of polygon.

$B C=A D$ which are opposite sides of polygon.

Also the diagonals $A C$ and $B D$ are not equal in length.

Hence, the polygon is not a rectangle.