Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of a square.

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Solution

Let A(3,3,3) , B(0,6,3) , C( 1,7,7) and D (4,4,7) are the vertices of quadrilateral $□ A B C D$

We have :

AB = $\sqrt{{(0 - 3)}^{2} + {(6 - 3)}^{2} + {(3 - 3)}^{2}}$
$= \sqrt{{(- 3)}^{2} + {(3)}^{2} + {(0)}^{2}} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3 \sqrt{2}$

BC = $\sqrt{{(1 - 0)}^{2} + {(7 - 6)}^{2} + {(7 - 3)}^{2}}$
$= \sqrt{{(1)}^{2} + {(1)}^{2} + {(4)}^{2}} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3 \sqrt{2}$

CD = $\sqrt{{(4 - 1)}^{2} + {(4 - 7)}^{2} + {(7 - 7)}^{2}}$
$= \sqrt{{(3)}^{2} + {(- 3)}^{2} + {(0)}^{2}} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3 \sqrt{2}$

DA = $\sqrt{{(4 - 3)}^{2} + {(4 - 3)}^{2} + {(7 - 3)}^{2}}$
$= \sqrt{{(1)}^{2} + {(1)}^{2} + {(4)}^{2}} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3 \sqrt{2}$

AB = BC = CD = DA

AC = $\sqrt{{(1 - 3)}^{2} + {(7 - 3)}^{2} + {(7 - 3)}^{2}}$
$= \sqrt{{(- 2)}^{2} + {(4)}^{2} + {(4)}^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$

BD = $\sqrt{{(4 - 0)}^{2} + {(4 - 6)}^{2} + {(7 - 3)}^{2}}$
$= \sqrt{{(4)}^{2} + {(- 2)}^{2} + {(4)}^{2}} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
$∴$ AC = BD
Since, all sides and diagonals of quadrilateral $□ A B C D$ are equal
Therefore, the points are the vertices of a square.

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