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(C) Let the position vectors be $\vec{A} = 7 \hat{i}-4 \hat{j}+7 \hat{k}$,$\vec{B} = \hat{i}-6 \hat{j}+10 \hat{k}$,$\vec{C} = -\hat{i}-3 \hat{j}+4 \ha

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a quadrilateral which is not a parallelogram

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(C) Let the position vectors be $\vec{A} = 7 \hat{i}-4 \hat{j}+7 \hat{k}$,$\vec{B} = \hat{i}-6 \hat{j}+10 \hat{k}$,$\vec{C} = -\hat{i}-3 \hat{j}+4 \hat{k}$,and $\vec{D} = 5 \hat{i}-\hat{j}+5 \hat{k}$.Calculating the side vectors:$\overrightarrow{AB} = \vec{B} - \vec{A} = (1-7)\hat{i} + (-6+4)\hat{j} + (10-7)\hat{k} = -6 \hat{i} - 2 \hat{j} + 3 \hat{k}$.$\overrightarrow{BC} = \vec{C} - \vec{B} = (-1-1)\hat{i} + (-3+6)\hat{j} + (4-10)\hat{k} = -2 \hat{i} + 3 \hat{j} - 6 \hat{k}$.$\overrightarrow{CD} = \vec{D} - \vec{C} = (5+1)\hat{i} + (-1+3)\hat{j} + (5-4)\hat{k} = 6 \hat{i} + 2 \hat{j} + \hat{k}$.$\overrightarrow{DA} = \vec{A} - \vec{D} = (7-5)\hat{i} + (-4+1)\hat{j} + (7-5)\hat{k} = 2 \hat{i} - 3 \hat{j} + 2 \hat{k}$.For a quadrilateral to be a parallelogram,opposite sides must be equal and parallel,i.e.,$\overrightarrow{AB} = \overrightarrow{DC}$ (or $\overrightarrow{AB} = -\overrightarrow{CD}$) and $\overrightarrow{BC} = \overrightarrow{AD}$ (or $\overrightarrow{BC} = -\overrightarrow{DA}$).Here,$\overrightarrow{AB} = -6 \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\overrightarrow{DC} = -\overrightarrow{CD} = -6 \hat{i} - 2 \hat{j} - \hat{k}$.Since $\overrightarrow{AB} 
eq \overrightarrow{DC}$,the quadrilateral is not a parallelogram.

If the position vectors of the vertices $A, B, C, D$ of a quadrilateral are $7 \hat{i}-4 \hat{j}+7 \hat{k}, \hat{i}-6 \hat{j}+10 \hat{k}, -\hat{i}-3 \hat{j}+4 \hat{k}$,and $5 \hat{i}-\hat{j}+5 \hat{k}$ respectively,then $ABCD$ is

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