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I'm asked to find an equation of the plane $(π)$ through the point $P(2,1,-1)$ which is perpendicular to the planes $(π_1):2x+y-3=0,(π_2):x+2y+z=2$

My try was: $\mathbf n_1 \times \mathbf n_2=(1,-2,3)$ is parallel to $(π)$ and $(π)$ is perpendicular to the line $(ε)$, where $(ε)=(π_1) \cap (π_2)$

We observe that $P(1,1,-1) \in (ε) \implies \mathbf x=(1,1,-1)+t(1,-2,3)$

From this point it's not very clear to me how we can find the equation of the plane if we know one point on the plane and a line that is perpendicular to it. Thanks in advance.

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    Please see [here](http://math.stackexchange.com/questions/663619/find-a-plane-that-passes-through-a-point-and-is-perpendicular-to-2-planes). – 2017-02-10

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You've correctly evaluated the cross product to be: $$\mathbf{n}_1\times \mathbf{n}_2=\hat{i}-2\hat{j}+3\hat{k}$$ Therefore, your plane should be of the following cartesian form: $$x-2y+3z=d$$ Evaluate $d$ by substituting $P(2,1,-1)$ into $x,y,z$.

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    Thank you. Do you know where I can find a visual representation of this for better understanding? – 2017-02-10
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    @Michael There are some nice diagrams on the web. Search for “normal equation of plane.” The basic idea is that the plane (more generally, line or hyperplane) consists of all points with the same projection onto the normal vector. – 2017-02-10