If A(-1, 2, 3),B(1, 1, 1) and C(2, -1, 3) are | vedclass

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(A) Given points are $A(-1, 2, 3)$,$B(1, 1, 1)$,and $C(2, -1, 3)$.First,we find two vectors lying on the plane:$\overrightarrow{AB} = (1 - (-1))i + (1

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$\pm \left( \frac{2i + 2j + k}{3} \right)$

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(A) Given points are $A(-1, 2, 3)$,$B(1, 1, 1)$,and $C(2, -1, 3)$.First,we find two vectors lying on the plane:$\overrightarrow{AB} = (1 - (-1))i + (1 - 2)j + (1 - 3)k = 2i - j - 2k$$\overrightarrow{AC} = (2 - (-1))i + (-1 - 2)j + (3 - 3)k = 3i - 3j + 0k$To find a normal vector to the plane,we calculate the cross product $\overrightarrow{AB} 	imes \overrightarrow{AC}$:$\overrightarrow{AB} 	imes \overrightarrow{AC} = \begin{vmatrix} i & j & k \ 2 & -1 & -2 \ 3 & -3 & 0 \end{vmatrix} = i(0 - 6) - j(0 - (-6)) + k(-6 - (-3)) = -6i - 6j - 3k$Let $\vec{n} = -6i - 6j - 3k$. The magnitude is $|\vec{n}| = \sqrt{(-6)^2 + (-6)^2 + (-3)^2} = \sqrt{36 + 36 + 9} = \sqrt{81} = 9$.The unit normal vector is $\hat{n} = \pm \frac{\vec{n}}{|\vec{n}|} = \pm \frac{-6i - 6j - 3k}{9} = \pm \left( \frac{-2i - 2j - k}{3} ight) = \pm \left( \frac{2i + 2j + k}{3} ight)$.

If $A(-1, 2, 3)$,$B(1, 1, 1)$ and $C(2, -1, 3)$ are points on a plane,then a unit normal vector to the plane $ABC$ is:

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