Let us draw the diagram first, Let L$$_{1}$$: 2x+ 3y -1 = 0 L$$_{2}$$: x+ 2y -3 = 0
L$$_{3}$$: 5x-6y- 1 = 0 With respect to L$$_{1}$$, we can see that the point $$(a, a^{2})$$ lies within the triangle and (0, 0) are opposite side. Therefore,  L$$_{(a, a^2)}$$* L$$_{(0, 0)}$$ < 0 $$Rightarrow$$ $$(2a+ 3a^2-1)(-1)0$$
$$Rightarrow$$ a < -1 or a > $$frac{1}{3}$$   ... (1)
With respect to L$$_{2}$$, we can see that the point $$(a, a^{2})$$ lies within the triangle and (0, 0) are on the same side. Therefore,  L$$_{(a, a^2)}$$* L$$_{(0, 0)}$$ > 0 $$Rightarrow$$ $$(a+ 2a^2-3)(-3)>0$$ $$Rightarrow$$ $$(2a^2+a-3) 0 $$Rightarrow$$ $$(5a- 6a^2-1)(-1)>0$$ $$Rightarrow$$ $$(6a^2-5a+1)>0$$ $$Rightarrow$$ a < $$frac{1}{3}$$ or a > $$frac{1}{2}$$   ... (3) From equation (1), (2) and (3) we can say that  a $$epsilon$$ (-3/2,-1) ⋃ (1/2, 1). Hence, option C is the correct answer.