How to move a point along a plane? So lets say Point P $(-3, -5, -1)$ is on a Plane $2x+2y-z=-15$. And lets say I want to move this point $3$ units left or $3$ units right along the plane, how can I do that?
I feel like doing this doesn't make sense:
$(-3 + 3, -5 + 3, -1 + 3)$
Like I don't want to move the point away from the plane, I want the point still be on the plane after the point moves.
 A: 
Got a hint from the diagram?

 All you need to do is find an equation of a circle with radius 3 lying on the plane $2x+2y-z=-15$ with the point $(-3,-5,-1)$ as the center and then all the points lying on the circle is the point 3 units away from $(-3,-5,-1)$.

Finding the locus of points 3 units away from Point P...

*

*The normal vector of the plane is $2\widehat i+2\widehat j-\widehat k$

*Find another point on the plane, so for simplicity substitute $x \text{ and } y =0$, we get $z=-15$. Another point on the plane is $(0,0,-15)$.

*Vector between P and $(0,0,-15)$ is $3\widehat i +5\widehat i -14\widehat k$

*Cross product between the normal vector ($2\widehat i+2\widehat j-\widehat k$) and $3\widehat i +5\widehat i -14\widehat k$ is $-23\widehat i -25\widehat j +4\widehat k$ whose direction ratios parallel to this will lie on the plane.

So a line passing through the point P is
$$\frac{x+3}{-23}=\frac{y+5}{-25}=\frac{z+1}{4}$$
Likewise, find another point on the plane & find the equation of the line passing through the Point P (probably this time you substitute $x \text{ and } z =0$). And using the equations of lines obtained, find the equation of the circle passing through them.
