Answer
$x-2=-2$
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:
$\hat{n}=\begin{vmatrix}i&j&k\\2&2&2\\-3&-5&-3\end{vmatrix}=4\hat{i}-4\hat{k}$
The plane containing the point $P(3,4,5)$ and having the normal vector $n=\lt 4,0,-4\gt$ is expressed algebraically by the equation as follows:
$4(x-3)+0(y-4)+(-4)(z-5)=0$
or, $4x-4z-12+20= 0$
Thus, $x-2=-2$
