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Find the equation of a plane that passes through point $P(1,5,1)$, and is perpendicular to the planes $2x+y-2z=2$ and $x+3z=4$

My only guess so far is that we can obtain the plane's normal vector using $2x+y-2z=2$ but I'm clueless on how to involve $x+3z=4$

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  • $\begingroup$ Write it as $x+0\cdot y+3z=4$. (You want to take the cross product of the normal vectors.) $\endgroup$ – David Mitra Feb 4 '14 at 13:54
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$n_1=(2,1,-2),n_2=(1,0,3)$ and $n=n_1×n_2$ then $$[(x,y,z)-(1,5,1)].n=0$$

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