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

The correct option is (-\infty ,\frac{-11}{5}).

To determine the set of all possible real values of a, let us analyze the given quadratic equation as a function:

f\left(x\right)=x^2+2ax+(3a+10)

Since the coefficient of x^2 is 1 (which is positive, 1>0), the graph of y=f\left(x\right) represents a parabola opening upwards.

We are given that the roots \alpha and \beta satisfy \alpha <1<\beta. This means that the number 1 lies strictly between the two roots of the quadratic equation.

For an upward-opening parabola, a real number k lies strictly between its roots if and only if the value of the function evaluated at k is strictly negative:

f\left(1\right)<0

Substituting x=1 into f\left(x\right):

f\left(1\right)={\left(1\right)}^2+2a\left(1\right)+3a+10

f\left(1\right)=1+2a+3a+10

f\left(1\right)=5a+11

Now, applying the condition f\left(1\right)<0:

5a+11<0

5a<-11

a<-\frac{11}{5}

In interval notation, the range of all possible values for a is:

a\in (-\infty ,-\frac{11}{5})