It is required to find the equation of a plane $Q$ formed by point $B\,(5,2,0)$ and the line (d) of parametric equation
$$ \begin{align} x&=-2t+1\\ y&=2t-2 \\ z&=t \end{align}$$
What is the easiest way to find the equation ? Answer is: $x-y+4z-3 = 0$
On
To get a plane in the form you want just solve linear equations:
If the plane is $\,\pi: ax+by+cz+d=0\,$ , then it must be" (with $\,L=$ the line):
$$\begin{align*}B\in\pi&\implies &\text{I}&\;\;\;&5a+2b+d=0\\ t=0\implies (1,-2,0)\in L\cap \pi&\implies &\text{II}&\;\;\;&a-2b+d=0\\ t=1\implies (-1,0,1)\in L\cap\pi&\implies &\text{III}&\;\;\;&-a+c+d=0\\ t=-1\implies (3,-4,-1)\in L\cap\pi&\implies &\text{IV}&\;\;\;&3a-4b-c+d=0\end{align*}$$
Now some easy, albeit boring and slightly annoying, basic algebra:
$$\begin{align*}\text{I - II}&:\;\;\;4(a+b)=0\implies a=-b\\ \text{II}&:\;\;\;a+2a+d=0\implies d=-3a\\ \text{III}&:\;\;\;-a+c-3a=0\implies c=4a\end{align*}$$
Well, now just choose a nice non-zero value for any of the variables, say $\,a=1\,$ , and you get
$$x-y+4z-3=0$$
Note that any other non-zero value for $\,a\,$ would have given the very same plane but with proportional coefficients.
I think the easiest way is to find the value of following determinant:
$$\begin{vmatrix} (x-5) & (y-2) & (z-0)\\ 4 & 4 & 0\\ -2 &2 & 1 \end{vmatrix}=0$$ Note that the point $(-1,-2,0)$ is a point lying on the line and the vector $\langle -2,2,1\rangle$ is the leading vector for the line.