unit vector of three dimention

The plane contains the vectors AB and AC

vector AB = (-1 , 4 , 8) - (1 , 2 , 3) = (-2 , 2 , 5)

vector AC = (5 , 1 , -2) - (1 , 2 , 3) = (4 , -1 , -5)

AB X AC (cross product)

=
│i j k│
│-2 2 5│
│4 -1 -5│

= -5i + 10j - 6k

This is the vector which is perpendicular to the plane.

And so, the unit vector is

= (-5i + 10j - 6k) / sqrt[(-5^2) + (10)^2 + (-6)^2]

= (-5/sqrt(161))i + (10/sqrt(161))j - (6/sqrt(161))k