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Find the equation of a plane which is perpendicular to the plane $$\pi_1\equiv x-3y-z+1=0,$$ parallel to a line $$l\equiv\frac{x - 2}{2} = \frac{y -3}{-3} = \frac{z}{1}$$ and goes through point $P = (-1, 1, 2)$.

All I know it that the normal vector of the given plane is $\overrightarrow{n_1}=(1,-3,-1)$ and the direction vector of the given line is $\overrightarrow{d_l}=(2,-3,1)$.

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The plane you are looking for contains the normal vector to $\pi_1$, and also contains the direction vector of $l$. As you point out, those are $\langle 1, -3, -1\rangle$ and $\langle 2, -3, 1\rangle$. Can you use that information to figure out a normal vector to the plane you are looking for, then put that together with the given information about $P$ to find your answer?

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  • $\begingroup$ When two planes are perpendicular their normal vectors are also perpendicular, right? $\endgroup$ – user155200 Jun 14 '14 at 18:25
  • $\begingroup$ That's correct. $\endgroup$ – rogerl Jun 14 '14 at 18:36

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