Answer
$x-3y=2$
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:
$\hat{n}=\begin{vmatrix}i&j&k\\-1&\dfrac{-1}{3}&-2\\2&\dfrac{2}{3}&-4\end{vmatrix}=\dfrac{8}{3}\hat{i}-8\hat{j}+0\hat{k}$
The plane containing the point $P(3,\dfrac{1}{3},-5)$ and having the normal vector $n=\lt \dfrac{8}{3},-8,0\gt$ is expressed algebraically by the equation as follows:
$ \dfrac{8}{3}(x-3)-8(y- \dfrac{1}{3})+0(z-(-5))=0$
or, $\dfrac{8}{3} x-8-8y+\dfrac{8}{3}=0$
Thus, $x-3y=2$
