I suppose it depends on how you define the variance on $x$ and $z$, but this question seems simple to me: yes.
If $P(x,y,z)$ is the set of all points $x, y, z$ such that $y=5$, it seems clear that y=5 is a plane.
The only reason I ask is because I was challenged in vector analysis that this indeed is not a plane but all points in $\Bbb{R}^3$. Almost everyone in there agreed with our professor (who I enjoy), so I call on others to please prove me wrong or right.
I graphed 'y=5 plane' on wolframalpha and it indeed was a plane to no surprise to me, but perhaps there is a way of defining it that would allow for all $\Bbb{R}^3$, who knows.