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.
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 $x,y,z$.