1. Let $P(x,y)$ be the moving point.

   Given: $PA^2+PB^2=14$ where the fixed points are $(1,2)$ and $(-2,1)$.

   So, $[(x-1)^2+(y-2)^2]+[(x+2)^2+(y-1)^2]=14$

2. Expand and simplify:

   $(x^2-2x+1)+(y^2-4y+4)+(x^2+4x+4)+(y^2-2y+1)=14$

   $2x^2+2y^2+2x-6y+10=14$

   $2x^2+2y^2+2x-6y-4=0$

   Divide by $2$: $x^2+y^2+x-3y-2=0$

   Hence the locus is $f(x,y)=x^2+y^2+x-3y-2=0$

3. Find intersection with the $x$-axis ($y=0$):

   $x^2+x-2=0$ $(x+2)(x-1)=0$

   So the points are $A(1,0),\\ B(-2,0)$

4. Find intersection with the $y$-axis ($x=0$):

   $y^2-3y-2=0$

   $y=\frac{3\pm \sqrt{9+8}}{2}=\frac{3\pm \sqrt{17}}{2}$

   So the points are $C\left(0,\frac{3+\sqrt{17}}{2}\right),\quad D\left(0,\frac{3-\sqrt{17}}{2}\right)$

5. Area of quadrilateral $ACBD$

   The quadrilateral has vertices on the axes:

   • $A(1,0)$ and $B(-2,0)$ on the $x$-axis
   • $C\left(0,\frac{3+\sqrt{17}}{2}\right)$ and $D\left(0,\frac{3-\sqrt{17}}{2}\right)$ on the $y$-axis

   Its diagonals are $AB$ and $CD$.

   Length of $AB$: $AB=1-(-2)=3$

   Length of $CD$: $CD=\frac{3+\sqrt{17}}{2}-\frac{3-\sqrt{17}}{2}=\sqrt{17}$

   Since $AB$ lies on the $x$-axis and $CD$ lies on the $y$-axis, the diagonals are perpendicular.

   Therefore, $\text{Area} = \frac{1}{2}(AB)(CD)=\frac{1}{2}\cdot 3\cdot \sqrt{17}=\frac{3\sqrt{17}}{2}$

6. Final answer: $\boxed{\frac{3\sqrt{17}}{2}}$

This matches Option B.