1. Let the three distinct real numbers in G.P. be $a=\frac{b}{r},\quad b=b,\quad c=br,$ where $r\neq 0$ is the common ratio.

Since the numbers are distinct, we must have $r\neq 1.$ Also, for real numbers in G.P., $r$ is real.

2. Given $a+b+c=xb.$ Substitute $a=\frac{b}{r}$ and $c=br$: $\frac{b}{r}+b+br=xb.$ Assuming $b\neq 0$ (in a G.P. of distinct real numbers, $b=0$ would force all terms to be $0$), divide by $b$: $\frac{1}{r}+1+r=x.$ So, $x=r+1+\frac{1}{r}.$

3. We now test which given values of $x$ are possible for real $r\neq 0,1$.

Multiply by $r$: $r^2+(1-x)r+1=0.$ For real $r$, the discriminant must be non-negative: $\Delta=(1-x)^2-4\ge 0.$ Thus, $|1-x|\ge 2.$ So either $1-x\ge 2 \Rightarrow x\le -1,$ or $1-x\le -2 \Rightarrow x\ge 3.$

Hence possible values of $x$ are $x\in (-\infty,-1]\cup[3,\infty).$ But note that $x=3$ occurs when $r=1$, which is not allowed because the terms must be distinct. Therefore, $x\in (-\infty,-1]\cup(3,\infty).$

4. Check the options:

• A: $2$. Since $2$ is not in $(-\infty,-1]\cup(3,\infty)$, it is not possible.
• B: $-3$. Possible.
• C: $4$. Possible.
• D: $-2$. Possible.

Let us verify quickly:

• For $x=-3$: $r+1+\frac{1}{r}=-3 \Rightarrow r^2+4r+1=0,$ which has real roots.
• For $x=4$: $r^2-3r+1=0,$ which has real roots.
• For $x=-2$: $r^2+3r+1=0,$ which has real roots.

Thus the only value $x$ cannot be is $2$.

5. Therefore, the correct option is $\boxed{A}.$