Find a point on a line equidistant from the coordinate plane and from the origin point I can't find the ways how to solve this problem:
On the straight line $$\frac{x+1}{1}=\frac{y-4}{-2}=\frac{z+1}{2}$$ find a point that is equidistant from the coordinate plane $Oxy$ and from the origin point $O(0; 0)$
Please, can you make me some hints or refer to the helpfull materials? Thank You!
 A: It seems like you want hints, so I'll try to write things in a not obvious way.  The line is a whole set of points, each defined by an X, Y and Z.  You should be able to write two equations, distance from the plane, and distance from the origin point in terms of that X, Y and Z.  So you'll want them to be equal, so you can set those two distances to be the same value.  So you should now have an equation based on X, Y and Z equal to another equation also defined by X, Y and Z.  You'll want to simply that to one variable (e.g. both equations defined by X, Y or Z) so that you can solve that one value.  Good thing you have a line, where X, Y and Z are defined in terms of one another.
Attempt at non-obvious explaination 2, the more explicit version:
Your goal is to find points (Xn, Yn, Zn) such that:
$\ ( Xn + 1 ) = ( Yn - 4) / -2 $
$\ ( Xn + 1 ) = ( Zn + 1) /  2 $
(These two equations being true ensure that this point is on the line.)
And "is equidistant from the coordinate plane Oxy and from the origin point O(0;0)"
The distance to the coordinate plane 0xy is $\ |Zn| $  The distance to the origin point O(0;0) is $\sqrt( (Xn-0)^2 + (Yn-0)^2 ) $  These two must be equal, so $\ |Zn| = sqrt( (Xn-0)^2 + (Yn-0)^2 ) $
You can rewrite the entire equations, either in terms of some new variable $\ t$ or in terms of either $\ Xn, Yn$, or $\ Zn$ 
Either way, you should end up with a quadratic function that identify the two solutions. 
A: Well, you should observe that the given plane passes through the origin, and so you need a point of the line whose projection on $Oxy$ is $O$. Take the perpendicular to $Oxy$ at $0$ and find where they cross.
A: There appears to be an error in the problem, as there are no such points.  
The distance of a point to the $x$-$y$ plane is simply the absolute value of its $z$-coordinate. Converting the implicit equations of the line into parametric form produces $(t-1,4-2t,2t-1)$, so after squaring both distances, we’re looking for solutions of $$(t-1)^2+(4-2t)^2+(2t-1)^2=(2t-1)^2.$$ The discriminant of this quadratic equation is $(-18)^2-4\cdot5\cdot17 = -16$, therefore it has no real roots.
