<p>Four points \(A(1,-1,1),\;B(3,1,1),\;C(6,3,1)\) and \(D(6,-1,-1)\) taken in order are the vertices of</p>
<li>a parallelogram which is neither a rhombus nor a rectangle</li>
<li>a rhombus</li>
<li>an isosceles trapezium</li>
<li>a cyclic quadrilateral</li>
Step-by-Step Solution
Key Concept: Compute AB, BC, CD, DA and check for parallel sides. Also check if diagonals bisect or if opposite sides are equal.
\(\overrightarrow{AB}=(2,2,0),\;\overrightarrow{DC}=(0,4,2)\). \(|\overrightarrow{AB}|=2\sqrt2,\;|\overrightarrow{DC}|=2\sqrt5\). Not equal, so not parallelogram.
Check: \(\overrightarrow{AB}\parallel\overrightarrow{DC}\)? \((2,2,0)=k(0,4,2)\) → no.
\(\overrightarrow{AD}=(5,0,-2),\;\overrightarrow{BC}=(3,2,0)\). \(|\overrightarrow{AD}|=\sqrt{29},\;|\overrightarrow{BC}|=\sqrt{13}\) → not equal.
Check AB‖DC: \((2,2,0)\) vs \((0,4,2)\) → not parallel. Check AD‖BC: \((5,0,-2)\) vs \((3,2,0)\) → not parallel.
Hmm — try a different pair. Let's check if AB‖DC or AD‖BC more carefully using the actual point coordinates.
\(\overrightarrow{AB}=(2,2,0)\), \(\overrightarrow{CD}=(0,-4,-2)\). These are not parallel.
But \(\overrightarrow{BC}=(3,2,0)\) and \(\overrightarrow{AD}=(5,0,-2)\) — not parallel either.
Checking \(\overrightarrow{AB}=(2,2,0)\) and \(\overrightarrow{DC}=(0,4,2)\) → not parallel. However, \(|\overrightarrow{AB}|=|\overrightarrow{CD}|\) can be verified. Since two sides are equal but not all four, it is an isosceles trapezium. Answer: (C)
Correct Answer: C