Caluculus minimum question

Find the point on the plane x – y + z = 1 that is closest to the point (6, 7, 9)

Let be f(x,y,z) = x - y + z - 1 = 0 our plane and the point A(6 , 7 , 9 ) ;

The minimum distance from the point to the plane is the normal from the point to the plane ;
The normal has equation :

(x - 6) / 1 = (y - 7) / (-1) = (z - 9) / 1 = t ;

and the parametric equations are :

x = 6 + t
y = 7 - t
z = 9 + t

Intersection with the plane gives t :

6 + t - 7 + t + 9 + t - 1 = 0 ==>. t = - 7/3 ;

and point A' ( 11/3 , 28/3 , 20/3 ) the closest to A