Answer
$2x+3y+z=4$
Work Step by Step
The plane containing the point $P(p,q,r)$ and having the normal vector $n=\lt a,b,c\gt$ is expressed algebraically by the equation as follows:
$a(x-p)+b(y-q)+c(z-r)=0$
Here, the normal vector $n$ can be calculated as follows:
$\hat{n}=\begin{vmatrix}i&j&k\\2&-2&2\\0&2&-6\end{vmatrix}=8\hat{i}+12\hat{j}+4\hat{k}$
The plane containing the point $P(2,0,0)$ and having the normal vector $n=\lt 8,12,4\gt$ is expressed algebraically by the equation as follows:
$8(x-2)+12(y-0)+4(z-0)=0$
or, $8x+12y+4z=16$
Thus, $2x+3y+z=4$
