Finding the Closest Point on a Plane to a Given Point

[Char. Limit]

Homework Statement

Find the point on the plane x-y+z=4 closest to the point (1,2,3).

d=\frac{2}{\sqrt{3}}

Homework Equations

Hmm...

The Attempt at a Solution

As you can see, I've already solved for the shortest distance. But, knowing the plane I'm on and the distance, how do I find the point that lies on that plane, at that distance from (1,2,3)? Help...

 

[Inferior89]

Embed the point into a plane with the same normal vector as the given plane.

The shortest distance from the origin to any plane ax +by +cz = d is
\frac{|\mathbf{n}|}{d} where \mathbf{n} is the normal vector to the plane.

Do this for both the given plane and your new plane from and calculate the difference.