$(a)\:Square\:\:\qquad\:(b)\:Rhombus\:\:\qquad\:(c)\:Rectangle\:\:\qquad\:(d)\:Parallelogram.$

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Let the points be $A(5,0,2),\:B(2,-6,0),\:C(4,-9,6),\:D(7,-3,8)$

$AB=\sqrt {9+36+4}=7$

$BC=\sqrt{4+9+36}=7$

$BD=\sqrt {9+36+4}=7$

$AD=\sqrt{ 4+9+36}=7$

$\overrightarrow {AB}=(-3,-6,-2)$

$\overrightarrow {AD}=(2,-3,6)$

$\overrightarrow {AB}.\overrightarrow {AD}=(-3,-6,-2).(2,-3,6)=-6+18-12=0$

Since all the four sides are equal and angle between adjacent sides is $90^{\circ}$,

$ABCD$ is a square.

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