Let $$\alpha$$ and $$\beta$$ be the roots of the equation $$x^{2}+2ax+\left(3a+10\right)=0$$ such that $$\alpha < 1 < \beta$$. Then the set of all possible values of $$a$$ is :

Given:

$$x^{2}+2ax+\left(3a+10\right)=0$$

Now $$\alpha$$ and $$\beta$$ be the roots given that $$\alpha < 1 < \beta$$.

Thus roots are distinct and real.

$$\Rightarrow$$ $$D > 0$$

$$\Rightarrow$$ $$4a^2 - 4(3a + 10) > 0$$

$$\Rightarrow$$ $$a^2 - 3a - 10 > 0$$

$$\Rightarrow$$ $$(a - 5)(a + 2) > 0$$

$$\Rightarrow$$ $$a < - 2$$ OR $$ a > 5$$   ----- (i)

The graph of the equation will be an upward parabola as the coefficient of $$x^2$$ is $$1$$, which is greater than $$0$$

Thus, we if we put $$x = 1$$ in the given quadratic equation, the value should be negative.

$$\Rightarrow$$ $$1 + 2a + (3a + 10) < 0$$

$$\Rightarrow$$ $$ 5a < -11$$

$$\Rightarrow$$ $$a < -\dfrac{11}{5}$$ ----(ii)

Thus by taking the intersection of (i) and (ii), we get the final interval for $$a$$ as,

$$a \in \left(-\infty, -\dfrac{11}{5}\right)$$

Hence, option A is the correct choice.