$$f(1) = a + b + c = 3$$ ..... (i)

$$f(3) = 9a + 3b + c = 4$$ .... (ii)

$$f(0) + f(1) + f( - 2) + f(3) = 14$$

OR $$c + 3 + (4a - 2b + c) + 4 = 14$$

OR $$4a - 2b + 2c = 7$$ ..... (iii)

From (i) and (ii) $$8a + 2b = 1$$ ..... (iv)

From (iii) $$ - (2) \times $$ (i)

$$ \Rightarrow 2a - 4b = 1$$ ..... (v)

From (iv) and (v) $$a = {1 \over 6},\,b = {{ - 1} \over 6}$$ and $$c = 3$$

$$f( - 2) = 4a - 2b + c$$

$$ = {4 \over 6} + {2 \over 6} + 3 = 4$$