Let #barA =# the vector from the first point to the second
#barA = (3 -1)hati + (-1 - 3)hatj + (2 -5)hatk#
#barA = 2hati - 4hatj - 3hatk#
Let #barB =# the vector from the first point to the third
#barB = (4 -1)hati + (0 - 3)hatj + (1 -5)hatk#
#barB = 3hati - 3hatj - 4hatk#
A vector perpendicular to the plane can be found by computing the cross product of these two vectors:
#AxxB =
|(hati, hatj, hatk, hati, hatj),(2, - 4, - 3, 2, -4)
,(3, -3, -4, 3, -3)
| =#
#hati{(-4)(-4) - (-3)(-3)} = 7hati#
#hatj{(-3)(3) - (2)(-4)} = -1hatj#
#hatk{(2)(-3) - (-4)(3)} = 6hatk#
A vector #barC# perpendicular to the plane is:
#barC = 7hati - hatj + 6 hatk#
However, you want the unit vector #hatC#
#hatC = barC/|barC|#
#|barC| = sqrt(7^2 + (-1)^2 + 6^2) #
#|barC| = sqrt(86) #
#hatC = 7sqrt(86)/86hati - sqrt(86)/86hatj + 6sqrt(86)/86hatk#
