## Introduction

In mathematics, the closest point problem is a problem of finding the point on a given set (usually a metric space) that is closest to a given point. It is a generalization of the problem of finding the closest point on a straight line or on a curve to a given point.

## Setting up the problem.

We are given a plane defined by the equation x + 2y + 3z = 6 and a point P(2, 7, 8). We want to find the point on the plane that is closest to P.

There are many ways to approach this problem, but one way is to set up a system of equations and solve for the unknowns. We can start by creating two vectors:

Vector v1: (x1, y1, z1) = P – Q

Vector v2: (x2, y2, z2) = n

where Q is an arbitrary point on the plane and n is the normal vector to the plane. We know that Q must satisfy the equation of the plane, so we can write:

x + 2y + 3z = 6 —-> x2 + 2y2 + 3z2 = 6

Now we have two equations and three unknowns (x1, y1, z1), so we can solve for x1 and y1 in terms of z1:

x1 = 2 – z1 —-> x2 = 2 – z2

y1 = 7 – 3z1 —-> y2 = 7 – 3z2

Substituting these values into our expression for v1 gives us:

v1 = (2 – z1, 7 – 3z1, z1) – (x2, y2, z2) —-> v1 = (-x2 + 2, -y2 + 7, -z2 + z1)

## Finding the point on the plane

There are many methods that can be used to find the point on a plane that is closest to a given point. One method is as follows:

1) Place the plane so that it intersects with the given point.

2) Find the point on the plane that is closest to the given point. This will be the point of intersection.

3) Measure the distance between the given point and the point of intersection. This will be the shortest distance between the two points.

## Conclusion

In conclusion, the point on the plane x y z 2 that is closest to the point 2 7 8 is the point (2, 7, 8).