Let R be the interior region between the lines $$3x - y + 1 = 0$$ and $$x + 2y - 5 = 0$$ containing the origin. The set of all values of $$a$$, for which the points $$(a^2, a + 1)$$ lie in R, is :

For the origin $$(0,0)$$ to be in the region, the signs of the expressions for $$(a^2, a+1)$$ must match the signs of $$(0,0)$$.

1. Line 1: $$L_1(0,0) = 1 > 0$$. So, $$3(a^2) - (a+1) + 1 > 0 \implies 3a^2 - a > 0 \implies a(3a-1) > 0$$.

Roots: $$0, 1/3$$. Intervals: $$(-\infty, 0) \cup (1/3, \infty)$$.

2. Line 2: $$L_2(0,0) = -5 < 0$$. So, $$(a^2) + 2(a+1) - 5 < 0 \implies a^2 + 2a - 3 < 0 \implies (a+3)(a-1) < 0$$.

Roots: $$-3, 1$$. Interval: $$(-3, 1)$$.

Intersection: $$(-3, 0) \cup (1/3, 1)$$