Let the three numbers in increasing GP be $a,\ ar,\ ar^2$ where, since they are increasing, we take $r>1$.

Condition after doubling the middle term

After doubling the middle number, the three numbers become: $a,\ 2ar,\ ar^2$ and these are in AP.

For three numbers in AP, the middle term is the average of the other two: $2(2ar)=a+ar^2$ $4ar=a+ar^2$

Dividing by $a$: $4r=1+r^2$ $r^2-4r+1=0$

Solving: $r=2\pm\sqrt{3}$

Since the GP is increasing, $r>1$, so $r=2+\sqrt{3}$

Use the fourth term condition

The GP is $a, ar, ar^2, ar^3$.

Given the fourth term is $3r^2$, so $ar^3=3r^2$ $a=\frac{3}{r}$

Find the common difference $d$ of the AP

The AP is: $a,\ 2ar,\ ar^2$

Hence common difference: $d=2ar-a=a(2r-1)$

Using $a=\frac{3}{r}$: $d=\frac{3}{r}(2r-1)=6-\frac{3}{r}$

Now rationalize $\frac{1}{r}$ for $r=2+\sqrt{3}$: $\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$

Therefore, $\frac{3}{r}=3(2-\sqrt{3})=6-3\sqrt{3}$

So, $d=6-(6-3\sqrt{3})=3\sqrt{3}$

Compute $r^2-d$

First, $r^2=(2+\sqrt{3})^2=4+3+4\sqrt{3}=7+4\sqrt{3}$

Hence, $r^2-d=(7+4\sqrt{3})-3\sqrt{3}=7+\sqrt{3}$

Match with options

$r^2-d=7+\sqrt{3}$

So the correct option is: B