A plane passes through (2,1,2) and (1,2,1) an | vedclass

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(C) The plane passes through $A(2,1,2)$ and $B(1,2,1)$. The vector $\vec{AB} = (1-2, 2-1, 1-2) = (-1, 1, -1)$.The line is given by $2x = 3y$ and $z =

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$(-2,0,1)$

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(C) The plane passes through $A(2,1,2)$ and $B(1,2,1)$. The vector $\vec{AB} = (1-2, 2-1, 1-2) = (-1, 1, -1)$.The line is given by $2x = 3y$ and $z = 1$. This can be written as $\frac{x}{3} = \frac{y}{2}$ and $z = 1$. The direction vector of the line is $\vec{v} = (3, 2, 0)$.The normal vector $\vec{n}$ to the plane is $\vec{AB} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & -1 \ 3 & 2 & 0 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(0 - (-3)) + \hat{k}(-2 - 3) = 2\hat{i} - 3\hat{j} - 5\hat{k}$.The equation of the plane is $2(x-2) - 3(y-1) - 5(z-2) = 0$,which simplifies to $2x - 4 - 3y + 3 - 5z + 10 = 0$,or $2x - 3y - 5z + 9 = 0$.Checking the options:For $(-6, 2, 0)$: $2(-6) - 3(2) - 5(0) + 9 = -12 - 6 + 9 = -9 
eq 0$.For $(6, -2, 0)$: $2(6) - 3(-2) - 5(0) + 9 = 12 + 6 + 9 = 27 
eq 0$.For $(-2, 0, 1)$: $2(-2) - 3(0) - 5(1) + 9 = -4 - 5 + 9 = 0$. Thus,the plane passes through $(-2, 0, 1)$.

$A$ plane passes through $(2,1,2)$ and $(1,2,1)$ and is parallel to the line $2x = 3y$ and $z = 1$. Then the plane also passes through which of the following points?

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