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(B) Let the points be $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$.First,we find two vectors in the plane: $\overrightarrow{PQ} = (2-1)i + (0-(-1))j +

Vedclass · Accepted Answer

$\frac{2i + j + k}{\sqrt{6}}$

Vedclass · Answer

(B) Let the points be $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$.First,we find two vectors in the plane: $\overrightarrow{PQ} = (2-1)i + (0-(-1))j + (-1-2)k = i + j - 3k$ and $\overrightarrow{PR} = (0-1)i + (2-(-1))j + (1-2)k = -i + 3j - k$.$A$ vector perpendicular to the plane is given by the cross product $\vec{n} = \overrightarrow{PQ} 	imes \overrightarrow{PR}$.$\vec{n} = \begin{vmatrix} i & j & k \ 1 & 1 & -3 \ -1 & 3 & -1 \end{vmatrix} = i(-1 - (-9)) - j(-1 - 3) + k(3 - (-1)) = 8i + 4j + 4k$.To simplify,we can take a parallel vector $\vec{v} = 2i + j + k$.The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.The unit vector perpendicular to the plane is $\pm \frac{\vec{v}}{|\vec{v}|} = \pm \frac{2i + j + k}{\sqrt{6}}$.Comparing this with the given options,the correct option is $\frac{2i + j + k}{\sqrt{6}}$.

$A$ unit vector perpendicular to the plane determined by the points $P(1, -1, 2)$,$Q(2, 0, -1)$,and $R(0, 2, 1)$ is:

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