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(D) Let the two given vertices be $\vec{A} = \hat{i} - \hat{j}$ and $\vec{B} = \hat{j} + \hat{k}$.For any point $\vec{C}$ to be the third vertex of a

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(D) Let the two given vertices be $\vec{A} = \hat{i} - \hat{j}$ and $\vec{B} = \hat{j} + \hat{k}$.For any point $\vec{C}$ to be the third vertex of a triangle,the points $\vec{A}$,$\vec{B}$,and $\vec{C}$ must not be collinear.Three points are collinear if the vectors $\vec{AB}$ and $\vec{AC}$ are parallel.Here,$\vec{AB} = \vec{B} - \vec{A} = (\hat{j} + \hat{k}) - (\hat{i} - \hat{j}) = -\hat{i} + 2\hat{j} + \hat{k}$.Checking option $(a)$: $\vec{C} = \hat{i} + \hat{k}$,then $\vec{AC} = (\hat{i} + \hat{k}) - (\hat{i} - \hat{j}) = \hat{j} + \hat{k}$. Since $\vec{AC}$ is not a scalar multiple of $\vec{AB}$,they are not collinear.Checking option $(b)$: $\vec{C} = \hat{i} - 2\hat{j} - \hat{k}$,then $\vec{AC} = (\hat{i} - 2\hat{j} - \hat{k}) - (\hat{i} - \hat{j}) = -\hat{j} - \hat{k}$. Since $\vec{AC}$ is not a scalar multiple of $\vec{AB}$,they are not collinear.Checking option $(c)$: $\vec{C} = \hat{i} - \hat{k}$,then $\vec{AC} = (\hat{i} - \hat{k}) - (\hat{i} - \hat{j}) = \hat{j} - \hat{k}$. Since $\vec{AC}$ is not a scalar multiple of $\vec{AB}$,they are not collinear.Since none of the given vectors are collinear with the side $\vec{AB}$,all of them can form a triangle with the given vertices.

If two vertices of a triangle are $\hat{i} - \hat{j}$ and $\hat{j} + \hat{k}$,then the third vertex can be

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