Show that the points `A(1,2,3),\ B(-1,-2,-1),\ C(2,3,2)a n d\ D(4,7,6)` are the vertices of a parallelogram ABCD but not a rectangle.

Explanation

Given that the point `A(1,2,3),\ B(-1,-2,-1),\ C(2,3,2)\ a n d\ D(4,7,6)`
we have to show that `A,B,C,D`are the vertices of a parallelogram and also ABCD is not a rectangle .
So ,`AB=sqrt((1+1)^2+(2+2)^2+(3+1)^2)`
`=sqrt(4+16+16)`
`=sqrt(36)`
`=6`
`CD=sqrt((2-4)^2+(3-7)^2+(2-6)^2)`
`=sqrt(4+16+16)`
`=sqrt(36)`
`=6`
`BC=sqrt((-1-2)^2+(-2-3)^2+(-1-2)^2)`
`=sqrt(9+25+9)`
`=sqrt(43)`
`DA=sqrt((4-1)^2+(7-2)^2+(6-3)^2)`
`=sqrt(9
+25+9)`
`=sqrt(43)`
It is clear from the above `AB=CD and BC=DA`
Hence it is a parallelogram.
Again we have to show It is not a rectangle.
`AC=sqrt((1-2)^2+(2-3)^2+(3-2)^2)`
`=sqrt(1+1+1)`
`=sqrt(3)`
`BD=sqrt((-1-4)^2+(-2-7)^2+(-1-6)^2)`
`=sqrt(25+81+49)`
`=sqrt(155)`
So, `ACneBD`
Therefore ABCD is not a rectangle.