Let the area of the region $$\left\{(x,y): 2y \leq x^{2}+3,y+|x| \leq 3,y \geq |x-1|\right\}$$ be A.Then 6 A is equal to :

Parabola: $$y = \frac{x^2+3}{2}$$ (Vertex at $$(0, 1.5)$$, opens up).

Line/V-shape: $$y = 3 - |x|$$.

Line/V-shape: $$y = |x - 1|$$.

By finding intersection points and integrating the "top curve minus bottom curve" over the bounded regions:

• Region 1 (left): $$\int_{x_1}^{x_2} ((3+x) - (\frac{x^2+3}{2})) dx$$

• Region 2 (right): $$\int_{x_2}^{x_3} ((3-x) - |x-1|) dx$$

Calculating the definite integral yields $$A = \frac{7}{3}$$.

Then $$6A = 6 \times \frac{7}{3} = \mathbf{14}$$.

Correct Option: C