A plane can be described by
This plane has normal vector 
.
. Suppose 
is a given external point. Find the point 
in the plane which is closest to 
.
is a given external point. Find the point in the plane which is closest to .The equation which I derived to solve this problem is
We can implement the equation above in Matlab, recalling that prime denotes transpose in Matlab.
In your example, the plane is
x+y+z=0
and therefore 
and D=0. Also, in your example, the external point is 
.
and D=0. Also, in your example, the external point is . The Matlab code to solve this is:
n=[1;1;1];
D=0;
p1=[1;1;1];
p0=p1-((n'*p1-D)/(n'*n))*n;
disp(p0)
Try it.
