Let a curve $$y = f(x)$$, $$x \in (0, \infty)$$ pass through the points $$P\left(1, \dfrac{3}{2}\right)$$ and $$Q\left(a, \dfrac{1}{2}\right)$$. If the tangent at any point $$R(b, f(b))$$ to the given curve cuts the y-axis at the point $$S(0, c)$$ such that $$bc = 3$$, then $$(PQ)^2$$ is equal to

Given a curve $$y = f(x)$$ passing through $$P\left(1, \frac{3}{2}\right)$$ and $$Q\left(a, \frac{1}{2}\right)$$, where the tangent at any point $$R(b, f(b))$$ cuts the y-axis at $$S(0, c)$$ with $$bc = 3$$.

The tangent at $$R(b, f(b))$$ is: $$Y - f(b) = f'(b)(X - b)$$

At $$X = 0$$: $$c = f(b) - b \cdot f'(b)$$

Since $$bc = 3$$ for all points on the curve (replacing $$b$$ with $$x$$):

$$x[f(x) - x f'(x)] = 3$$

$$xf(x) - x^2 f'(x) = 3$$

Rearranging:

$$f'(x) - \frac{f(x)}{x} = -\frac{3}{x^2}$$

This is a linear ODE with integrating factor $$e^{-\int \frac{1}{x}dx} = \frac{1}{x}$$:

$$\frac{d}{dx}\left[\frac{f(x)}{x}\right] = -\frac{3}{x^3}$$

Integrating:

$$\frac{f(x)}{x} = \frac{3}{2x^2} + C$$

$$f(x) = \frac{3}{2x} + Cx$$

Using $$f(1) = \frac{3}{2}$$:

$$\frac{3}{2} = \frac{3}{2} + C \implies C = 0$$

So $$f(x) = \frac{3}{2x}$$.

Finding $$a$$: $$f(a) = \frac{1}{2} \implies \frac{3}{2a} = \frac{1}{2} \implies a = 3$$

Therefore $$Q = (3, \frac{1}{2})$$ and:

$$(PQ)^2 = (3-1)^2 + \left(\frac{1}{2} - \frac{3}{2}\right)^2 = 4 + 1 = 5$$

The answer is $$5$$.