Let $$f(x) = ax^2 + bx + c$$ be such that $$f(1) = 3, f(-2) = \lambda$$ and $$f(3) = 4$$. If $$f(0) + f(1) + f(-2) + f(3) = 14$$, then $$\lambda$$ is equal to

Given,

$$f(x)=ax^2+bx+c$$

Also,

$$f(1)=3,\qquad f(-2)=\lambda,\qquad f(3)=4$$

and

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

Since

$$f(0)=c,$$

we get

$$c+3+\lambda+4=14$$

$$c+\lambda=7$$

$$c=7-\lambda$$

Now,

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

$$a+b+7-\lambda=3$$

$$a+b=\lambda-4$$

Also,

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

$$9a+3b+7-\lambda=4$$

$$9a+3b=\lambda-3$$

Dividing by $$3,$$

$$3a+b=\frac{\lambda-3}{3}$$

Subtracting,

$$2a=\frac{\lambda-3}{3}-(\lambda-4)$$

$$2a=\frac{\lambda-3-3\lambda+12}{3}$$

$$2a=\frac{9-2\lambda}{3}$$

$$a=\frac{9-2\lambda}{6}$$

Now use

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

Substituting

$$c=7-\lambda$$

and

$$b=\lambda-4-a,$$

we get

$$4a-2(\lambda-4-a)+7-\lambda=\lambda$$

$$6a-3\lambda+15=\lambda$$

$$6a=4\lambda-15$$

But,

$$a=\frac{9-2\lambda}{6}$$

Hence,

$$9-2\lambda=4\lambda-15$$

$$24=6\lambda$$

$$\lambda=4$$

Hence, $$\boxed{4}$$.