Answer
$y+z=2$
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:
$\hat{n}=\begin{vmatrix}i&j&k\\2&2&-2\\0&2&-2\end{vmatrix}=4\hat{j}+4\hat{k}$
The plane containing the point $P(\dfrac{3}{2},4,-2)$ and having the normal vector $n=\lt 0,4,4\gt$ is expressed algebraically by the equation as follows:
$0(x- \dfrac{3}{2})+4(y-4)+4(z-(-2))=0$
or, $4y+4z-16+8=0$
Thus, $y+z=2$
