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A plane through the origin is perpendicular to the plane $2x-y-z=5$ and parallel to the line joining the points $(1,2,3)$ and $(4,-1,2)$. Find the equation of the plane.

Analyzing this problem I found that the normal vector of the required plane is perpendicular to the line parallel to the plane. And, the given plane's normal vector is parallel to the line. I am stuck after this. Can I do a cross product between the vector formed by the two points and the normal vector of the given plane?

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We know that the required plane's normal is perpendicular to both the given plane's normal $(2,-1,-1)$ and the line's direction vector $(3,-3,-1)$. We can indeed perform a cross product to get the required plane's normal, since its result is perpendicular to both its inputs: $$(2,-1,-1)\times(3,-3,1)=\dots$$

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  • $\begingroup$ Aren't the direction vector of the line and the normal vector of the plane parallel to each other? and taking the cross product of the two of them results to 0? or am I missing something? $\endgroup$ – Joshua Roberto Gruta Mar 17 at 7:33
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    $\begingroup$ @JoshuaRobertoGruta It doesn't matter here. What is important is that both those vectors are perpendicular to the required plane's normal. $\endgroup$ – Parcly Taxel Mar 17 at 8:23
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You could as well find the parametric form of the plane: $$x=\lambda\begin{pmatrix}2\\-1\\-1\end{pmatrix}+\mu\begin{pmatrix}3\\-3\\1\end{pmatrix},\lambda,\mu\in\mathbb{R}$$ where the first direction vector ( the normal vector of the plane) indicates this plane is perpendicular to the given one and the second is the direction vector of the given line to which it is parallel and thus as pointed out by Parcly Taxel a normal vector of the desired plane is given by $$\begin{pmatrix}2\\-1\\-1\end{pmatrix}\times\begin{pmatrix}3\\-3\\1\end{pmatrix}=\begin{pmatrix}-4\\-5\\-3\end{pmatrix}$$ from what You get the coordinate equation of the plane: $$4x+5y+3z=0.$$

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