Parallelogram ABCD A(0,1,2)B(1,2,3)C(0,3,5)?

Let the coordinates of vertex D be (x,y,z)
Vector AB = Vector CD
(1-0)i + (2-1)j + (3-2)k = (x-0)i + (y-3)j + (z-5)k
xi + yj + zk = 1i + 4j + 6k
Coordinates of vertex D : (1,4,6)
Let angle of B be θ.
Vector BA DOT Vector BC = |BA||BC|cosθ
(1-0)(0-1) + (2-1)(3-2) + (3-2)(5-3) = √(1^2+1^2+1^2)√(1^2+1^2+2^2)cosθ
2 = √(18)cosθ
θ = 61.87 degree