Calculate the equation of the plane $\pi$ crossing a given point $P(1,-1,2) \in \pi$ and such that $\pi \parallel \pi_1$, where $\pi_1: 3x-y+2z=0$. Present the normal (general) form and the parameter form.
Here's what I have so far:
$$\vec{n_1}=\begin{pmatrix}3\\ -1\\ 2\end{pmatrix},\:\vec{n}=k\cdot \begin{pmatrix}3\\ \:-1\\ \:2\end{pmatrix}$$
$$k*3*1+k*(-1)*(-1)+k*2*(-2)+d=0 \implies d = -8k$$
$$a = 3k \\ b = -k \\ c = 2k \\ d = -8k$$
General form: $$ \pi:3k*x+(-k)*y+2k*z-8k=0 $$
Is that correct? How do I get the parameter form from there?
The plane must have the form $3x-y+2z+d=0.$ Put in the coordinates of the given point $P$ and solve for $d.$ To find the parametric form, let $x=s,z=t$ and solve for $y.$