We want to find the set of vectors parallel to the plane that are
perpendicular to the vector <1,2,1>.

Vectors in or parallel to the plane are vectors < x,y,z > such that 

x+2y+z = 0

Vectors perpendicular to <1,3,1> are < x,y,z > such that

<1,3,1>∙< x,y,z > = x+3y+z = 0

So we have this under-determined system



Subtracting them we get y = 0
Substituting we get x+2(0)+z = 0, or x+z=0, or z=-x

So the orthogonal complement is the set of vectors of the form

< k,0,-k >

That is the perpendicular distance from the point. We use the
formula for the perpendicular distance from the point
 
(x1, y1, z1) to the plane

ax + by + cz + d = 0, which is:

is 

So in our case (x1, y1, z1) = (3,4,1),
a=1, b=3, c=1, d=0







Edwin