The points A and B have position vectors, relative to the origin O, given by OA = i+ 2j + 3k and OB = 2i +j+ 3k. The line I has vector equation r = (1 – 21)i + (5 + t)j + (2 – t)k. (i) Show that I does not intersect the line passing through A and B. (ii) The point P lies on l and is such that angle PAB is equal to 60°. Given that the position vector ofP is (1- 2t)i + (5 + t)j + (2 – t)k, show that 3t2 + 7t+2 0. Hence find the only possible position vector of P.

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