A point $$P$$ moves so that the sum of squares of its distances from the points $$(1, 2)$$ and $$(-2, 1)$$ is $$14$$. Let $$f(x, y) = 0$$ be the locus of $$P$$, which intersects the $$x$$-axis at the points $$A, B$$ and the $$y$$-axis at the point $$C, D$$. Then the area of the quadrilateral $$ACBD$$ is equal to

A point P(x, y) moves such that the sum of squares of its distances from (1, 2) and (-2, 1) is 14.

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

Expanding each term:

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

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

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

Rewriting in standard form by completing the square:

$$\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + \left(y - \frac{3}{2}\right)^2 - \frac{9}{4} - 2 = 0$$

$$\left(x + \frac{1}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 = \frac{1}{4} + \frac{9}{4} + 2 = \frac{10}{4} + 2 = \frac{9}{2}$$

This is a circle with centre $$\left(-\frac{1}{2}, \frac{3}{2}\right)$$ and radius $$r = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$$.

Setting $$y = 0$$: $$x^2 + x - 2 = 0 \implies (x+2)(x-1) = 0$$

Points: $$A(-2, 0)$$ and $$B(1, 0)$$. So $$|AB| = 3$$.

Setting $$x = 0$$: $$y^2 - 3y - 2 = 0$$

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

Points: $$C\left(0, \frac{3+\sqrt{17}}{2}\right)$$ and $$D\left(0, \frac{3-\sqrt{17}}{2}\right)$$. So $$|CD| = \sqrt{17}$$.

The quadrilateral ACBD has its diagonals along the x-axis (AB) and y-axis (CD), which are perpendicular to each other. For a quadrilateral with perpendicular diagonals:

$$\text{Area} = \frac{1}{2} \times |AB| \times |CD| = \frac{1}{2} \times 3 \times \sqrt{17} = \frac{3\sqrt{17}}{2}$$

Therefore, the correct answer is Option B: $$\dfrac{3\sqrt{17}}{2}$$.