find the value of $a$;

determine the coordinates of the point of intersection P.

METHOD 1

${l_1}:$r $= \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ a \end{array}} \right) = \beta \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right) \Rightarrow \left\{ {\begin{array}{*{20}{l}} {x = - 3 + \beta } \\ {y = - 2 + 4\beta } \\ {z = a + 2\beta } \end{array}} \right.$     M1

$\frac{{6 - ( - 3 + \beta )}}{3} = \frac{{( - 2 + 4\beta ) - 2}}{4} \Rightarrow 4 = \frac{{4\beta }}{3} \Rightarrow \beta  = 3$    M1A1

$\frac{{6 - ( - 3 + \beta )}}{3} = 1 - (a + 2\beta ) \Rightarrow 2 =  - 5 - a \Rightarrow a =  - 7$    A1

METHOD 2

$\left\{ {\begin{array}{*{20}{l}} { - 3 + \beta = 6 - 3\lambda } \\ { - 2 + 4\beta = 4\lambda + 2} \\ {a + 2\beta = 1 - \lambda } \end{array}} \right.$    M1

attempt to solve     M1

$\lambda  = 2,{\text{ }}\beta  = 3$    A1

$a = 1 - \lambda  - 2\beta  =  - 7$    A1

[4 marks]

$\overrightarrow {{\text{OP}}} = \left( {\begin{array}{*{20}{c}} { - 3} \\ { - 2} \\ { - 7} \end{array}} \right) + 3 \bullet \left( {\begin{array}{*{20}{c}} 1 \\ 4 \\ 2 \end{array}} \right)$    (M1)

$= \left( {\begin{array}{*{20}{c}} 0 \\ {10} \\ { - 1} \end{array}} \right)$    A1

$\therefore {\text{P}}(0,{\text{ 10, }} - 1)$

[2 marks]