a)
B-A = (3, -3, -6)
So L: (0, 5, 5) + t(3, -3, -6)
[or (3, 2, -1) + t(3, -3, -6)]
b)
Since C lies on L, then it has position vector:
(3t, 5-3t, 5-6t)
OC . direction vector of L = 0
(3t, 5-3t, 5-6t) . (3, -3, -6) = 0
9t - 3(5-3t) - 6(5-6t) = 0
9t - 15 + 9t - 30 + 36t = 0 => t=5/6
Substituting in (3t, 5-3t, 5-6t) gives:
(5/2, 5/2, 0)
So position vector of C is:
2.5i + 2.5j
I've forgotten most of P3 vectors, so I'm not very sure my answer's right.