There may be an easier way, but this is a way I know will work.
The vectors AD and BC are both in the same plane, so that plane will have a normal that is in the direction of their cross product.
... AD × BC = [10 -6 4], which can be reduced to [5 -3 2]
Since AP ⊥ BC also lies in the same plane, it will be orthogonal to [5 -3 2] and to [2 4 1], so its direction can be found from the cross product of these vectors.
... [5 -3 2] × [2 4 1] = [-11 -1 26]
Now all we need to do is to find the factor this needs to be multiplied by to get the desired magnitude.
||[-11 -1 26]|| = √798, so our multiplying factor is (√6)/3/√798 = (√133)/399
Then the coordinates of P are [-11√133/399 -√133/399 26√133/399]
