A unit vector perpendicular to the plane determined by the points \[P\,(1,\,\,-1,\,\,2),\,\,Q\,(2,\,\,0,\,-1)\] and \[R\,(0,\,\,2,\,\,1)\] is [IIT 1994]

\[\frac{-2\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{6}}\]

\[\frac{2\mathbf{i}-\mathbf{j}+\mathbf{k}}{\sqrt{6}}\]

\[\frac{2\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{6}}\]

\[\frac{2\mathbf{i}+\mathbf{j}-\mathbf{k}}{\sqrt{6}}\]

A unit vector perpendicular to the plane determined by the points \[P\

1.	A unit vector perpendicular to the plane determined by the points \[P\,(1,\,\,-1,\,\,2),\,\,Q\,(2,\,\,0,\,-1)\] and \[R\,(0,\,\,2,\,\,1)\] is [IIT 1994]
A.	\[\frac{2\mathbf{i}-\mathbf{j}+\mathbf{k}}{\sqrt{6}}\]
B.	\[\frac{2\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{6}}\]
C.	\[\frac{-2\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{6}}\]
D.	\[\frac{2\mathbf{i}+\mathbf{j}-\mathbf{k}}{\sqrt{6}}\]
Answer» C. \[\frac{-2\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{6}}\]

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A unit vector perpendicular to the plane determined by the points \[P\,(1,\,\,-1,\,\,2),\,\,Q\,(2,\,\,0,\,-1)\] and \[R\,(0,\,\,2,\,\,1)\] is [IIT 1994]

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