10 The points \(A\) and \(B\) have position vectors, relative to the origin \(O\), given by $$\overrightarrow { O A } = \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } \quad \text { and } \quad \overrightarrow { O B } = 2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k } .$$ The line \(l\) has vector equation $$\mathbf { r } = ( 1 - 2 t ) \mathbf { i } + ( 5 + t ) \mathbf { j } + ( 2 - t ) \mathbf { k }$$

1. Show that \(l\) does not intersect the line passing through \(A\) and \(B\).
2. The point \(P\) lies on \(l\) and is such that angle \(P A B\) is equal to \(60 ^ { \circ }\). Given that the position vector of \(P\) is \(( 1 - 2 t ) \mathbf { i } + ( 5 + t ) \mathbf { j } + ( 2 - t ) \mathbf { k }\), show that \(3 t ^ { 2 } + 7 t + 2 = 0\). Hence find the only possible position vector of \(P\). \footnotetext{Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. }