Given the vertices a (1, - 1,2), B (5, - 6,2) C (1,3, - 1) of △ ABC, then the length of high BD on the edge of AC is (vector)

Given the vertices a (1, - 1,2), B (5, - 6,2) C (1,3, - 1) of △ ABC, then the length of high BD on the edge of AC is (vector)

Let D (x, y, z), due to the equation of AC
(x-1)/0=(y+1)/4=(z-2)/-3=t
The parameter equation is x = 1, y = 4t-1, z = - 3T + 2, so D (1,4t-1, - 3T + 2)
Because vector AC = (0,4, - 3)
BD=(x-5,y+6,z-2)=(-4,4t+5,-3t)
AC perpendicular to BD
therefore AC.BD=0
4*(4t+5)-3*(-3t)=0
16t+9t+20=0
t=-4/5
So BD = (- 4,9 / 5,12 / 5)
The length of high BD is | BD | = (16 + 81 / 25 + 144 / 25) ^ 1 / 2 = (16 + 225 / 25) ^ 1 / 2 = (16 + 9) ^ 1 / 2 = 5