The y-axis and the lines  $(a^{5 }– 2a^3 )x + (a + 2)y + 3a = 0$  and$(a^5 – 3a^2 )x + 4y + a – 2 = 0$ are concurrent for 

Let the lines intersect at the point k then 

$(a+2)k+3a=0$ and $4k+a-2=0$ 

$\Rightarrow \frac{3a}{a+2}=\frac{a-2}{4}\Rightarrow a^2-12a+4=0$

which gives us two values of a