The distance between the point and plane

How would I find the distance between the point (2,-3,1) and the plane 3x+y-2z=15
The distance from P(x,y,z) to the plane ax+by+cz=d is:
|ax+by+cz-d| / sqrt(a^2 + b^2 + c^2)
Therefore |ax+by+cz-d| = |3(2) + (-3) + (-2) - 15| = |-14| = 14.
sqrt(a^2 + b^2 + c^2) = sqrt (9 + 1 + 4) = sqrt14.
The distance is 14/sqrt14 or sqrt 14.