Please help with this math question: Vectors, 10 points to best

The position vectors A, B, C are given by: (i + j + k), (2i + 2j - 3k), (i - j + 2k) respectively.
AB = i + j - 4k, and AC= -2j + k:

Let L be the staight line passing through A and C. The perpendicular from B to the line L meet L at D. Find the length AD and hence find the perpendicular distance of B from L.

Please help, thanks a lot,10 points to best.

OA = i + j + k
OB = 2i + 2j - 3k
OC = i - j + 2k
AB = OB - OA = i + j - 4k
AC = OC - OA = -2j + k

The line passing through A and C is OA + t AC = i + (1 - 2t)j + (1 + t)k

So for some particular t, OD = i + (1 - 2t)j + (1 + t)k

AC must be perpendicular to BD
BD = OD - OB = i + (1 - 2t)j + (1 + t)k - (2i + 2j - 3k)
BD = (-1) i + (-1 - 2t) j + (4 + t) k

AC o BD = 0
(-2j + k) o ((-1) i + (-1 - 2t) j + (4 + t) k) = 0
0 + (2 + 4t) + (4 + t) = 0
5t + 6 = 0
t = -6/5

OD = i + (1 - 2t)j + (1 + t)k = i + (17/5) j - (1/5)k

AD = OD - OA = ( i + (17/5) j - (1/5)k) - ( i + j + k)
AD = (12/5) j - (6/5)k

AD = sqrt(144/25 + 36/25) = 6 / sqrt(5)

Points are:
A(1,1,1)
B(2,2,-3)
C(1,-1,2)

Vectors:
AB = i + j - 4k=(1,1,-4) and |AB|=√(1²+1²+(-4)²)=3√2
AC= -2j + k=(0,-2,1) and |AC|=√(0²+(-2)²+1²)=√5

AD=|(AB,AC)|/|AC|=|1*0+1*(-2)+(-4)*1|/…

BD=√(|AB|²-AD²)=√((3√2)²-(6/√5)²)=...