Question 23633
  the perpendicular distance from A(2,3,-1) to the plane 2x + y - 2z + 9 = 0.
 
 The distance formula of a point A(xo,y,zo) to the plane ax + b y + cz+d = 0

 is D = |axo + byo + czo + d|/ sqrt(a^2+b^2+c^2)...(**)

 where N = ai + bj + ck (=(a,b,c))is normal to the plane and 
 |N| = sqrt(a^2+b^2+c^2).

 The formula can be obtained by assuming the point P=(x,y,z) of A on the 
 plane and solve (x-xo,y-yo,z-zo) // (a,b,c) , ax+by+cz+d = 0 and
 let k = (x-xo,y-yo,z-zo) /(a,b,c), (x,y,z)= (xo+ka,yo+kb,zo+kc).
 a(xo+ka)+ b(yo+kb)+c(zo+kc)+d = 0, get k(a^2+b2+c^2) = -(axo+byo+czo+d). 
 D = |(x-xo,y-yo,z-zo)|  = |k| |(a,b,c)| = 
 |axo + byo + czo + d|/ sqrt(a^2+b^2+c^2) 

 It is the same value as your projection of AC to N.

 Kenny