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(B) To find a vector perpendicular to the plane containing points $A, B$,and $C$,we calculate the cross product of two vectors lying in the plane,such

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$8 \hat{i} + 4 \hat{j} + 4 \hat{k}$

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(B) To find a vector perpendicular to the plane containing points $A, B$,and $C$,we calculate the cross product of two vectors lying in the plane,such as $\vec{AB}$ and $\vec{AC}$.Given points: $A(1, -1, 2)$,$B(2, 0, -1)$,$C(0, 2, 1)$.$\vec{AB} = (2-1)\hat{i} + (0 - (-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + \hat{j} - 3\hat{k}$$\vec{AC} = (0-1)\hat{i} + (2 - (-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$The normal vector $\vec{n} = \vec{AB} 	imes \vec{AC} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -3 \ -1 & 3 & -1 \end{array}ight|$$\vec{n} = \hat{i}(-1 - (-9)) - \hat{j}(-1 - 3) + \hat{k}(3 - (-1))$$\vec{n} = \hat{i}(8) - \hat{j}(-4) + \hat{k}(4) = 8\hat{i} + 4\hat{j} + 4\hat{k}$Thus,the vector perpendicular to the plane is $8\hat{i} + 4\hat{j} + 4\hat{k}$.

$A$ vector perpendicular to the plane containing the points $A(1, -1, 2)$,$B(2, 0, -1)$,and $C(0, 2, 1)$ is

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