If the length of the latus rectum of the ellipse $$x^2 + 4y^2 + 2x + 8y - \lambda = 0$$ is $$4$$, and $$l$$ is the length of its major axis, then $$\lambda + l$$ is equal to_______.

Let $$f(x) = ax^2 + bx + c$$. We are given $$f(1) = 3$$, $$f(-2) = \lambda$$, $$f(3) = 4$$, and $$f(0) + f(1) + f(-2) + f(3) = 14$$.

From $$f(0) = c$$, the sum condition gives $$c + 3 + \lambda + 4 = 14$$, so $$c + \lambda = 7$$ $$-(1)$$.

From $$f(1) = a + b + c = 3$$ $$-(2)$$.

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

Subtracting $$(2)$$ from $$(3)$$: $$8a + 2b = 1$$ $$-(4)$$.

From $$f(-2) = 4a - 2b + c = \lambda$$ $$-(5)$$.

From $$(1)$$: $$c = 7 - \lambda$$. Substituting into $$(2)$$: $$a + b = 3 - (7 - \lambda) = \lambda - 4$$ $$-(6)$$.

Substituting $$c = 7 - \lambda$$ into $$(5)$$: $$4a - 2b + 7 - \lambda = \lambda$$, so $$4a - 2b = 2\lambda - 7$$ $$-(7)$$.

From $$(6)$$: $$b = \lambda - 4 - a$$. Substituting into $$(4)$$: $$8a + 2(\lambda - 4 - a) = 1$$, so $$6a + 2\lambda - 8 = 1$$, giving $$a = \dfrac{9 - 2\lambda}{6}$$ $$-(8)$$.

From $$(6)$$: $$b = \lambda - 4 - \dfrac{9 - 2\lambda}{6} = \dfrac{6\lambda - 24 - 9 + 2\lambda}{6} = \dfrac{8\lambda - 33}{6}$$.

Substituting into $$(7)$$: $$4 \cdot \dfrac{9-2\lambda}{6} - 2 \cdot \dfrac{8\lambda-33}{6} = 2\lambda - 7$$.

$$\dfrac{36 - 8\lambda - 16\lambda + 66}{6} = 2\lambda - 7$$

$$\dfrac{102 - 24\lambda}{6} = 2\lambda - 7$$

$$102 - 24\lambda = 12\lambda - 42$$

$$144 = 36\lambda$$, so $$\lambda = 4$$.

The answer is $$4$$.